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32*b 


IN  MEMORIAM 
FLORIAN  CAJORI 


4^i^^a^     (Sy^ 


WENTWORTH'S  "^ 

PLANE  GEOMETRY 


REVISED  BY 

GEORGE  WENTWORTH 

AND 

DAVID  EUGENE  SMITH 


GINN  AND  COMPANY 

BOSTON  ■  NEW  YORK  •  CHICAGO  •  LONDON 


COPYRIGHT,  1888,  1899,  BY  G.  A.  WENT  WORTH 

COPYRIGHT,  1910,   BY  GEORGE  WENTWORTH 

AND    DAVID   EUGENE   SMITH 

ENTERED   AT   STATIONERS'    HALL 

ALL  RIGHTS  RESERVED 

910.7 


GINN  AND  COMPANY  •  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


b 


^ 


I^Ai/^ 


PEEFACE 


Long  after  the  death  of  Robert  Recorde,  England's  first 
great  writer  of  textbooks,  the  preface  of  a  new  edition  of 
one  of  his  works  contained  the  appreciative  statement  that 
the  book  was  "  entail'd  upon  the  People,  ratified  and  sign'd  by 
the  approbation  of  Time.''  The  language  of  this  sentiment 
sounds  quaint,  but  the  noble  tribute  is  as  impressive  to-day 
as  when  first  put  in  print  two  hundred  fifty  years  ago. 

With  equal  truth  these  words  may  be  applied  to  the  Geome> 
try  written  by  George  A.  Wentworth.  For  a  generation  it 
has  been  the  leading  textbook  on  the  subject  in  America.  It 
set  a  standard  for  usability  that  every  subsequent  writer  upon 
geometry  has  tried  to  follow,  and  the  number  of  pupils  who 
have  testified  to  its  excellence  has  run  well  into  the  millions. 

In  undertaking  to  prepare  a  revision  of  this  work,  the  authors 
have  been  guided  by  certain  well-defined  principles,  based  upon 
an  extended  investigation  of  the  needs  of  the  schools  and  upon 
a  study  of  all  that  is  best  in  the  recent  literature  of  the  sub- 
ject. The  effects  of  these  principles  they  feel  should  be  sum- 
marized for  the  purpose  of  calling  the  attention  of  the  wide 
circle  of  friends  of  the  Wentworth  text  to  the  points  of  simi- 
larity and  of  difference  in  the  editions. 

1.  Every  effort  has. been  made  not  only  to  preserve  but  to 
improve  upon  the  simplicity  of  treatment,  the  clearness  of 
expression,  and  the  symmetry  of  page  that  have  characterized 
the  successive  editions  of  the  Wentworth  Geometry.  It  has 
been  the  purpose  to  prepare  a  book  that  should  do  even  more 
than  maintain  the  traditions  this  work  has  fostered. 


rft\£y^  €\9% 


iv  PLANE  GEOMETRY 

2.  The  proofs  have  been  given  substantially  in  full,  tOv.the 
end  that  the  pupil  may  always  have  before  him  a  model  for 
his  independent  treatment  of  the  exercises. 

3.  The  sequence  of  propositions  has  been  improved  in  sev- 
eral respects,  notably  in  the  treatment  of  parallels. 

4.  To  meet  a  general  demand,  the  number  of  propositions 
has  been  decreased  so  as  to  include  only  the  great  basal  theo- 
rems and  problems.  A  little  of  the  less  important  material 
has  been  placed  in  the  Appendix,  to  be  used  or  not  as  circum- 
stances demand. 

5.  The  exercises,  in  some  respects  the  most  important  part 
of  a  course  in  geometry,  have  been  rendered  more  dignified  in 
appearance  and  have  been  improved  in  content.  The  number 
of  simple  exercises  has  been  greatly  increased,  while  the  diffi- 
cult puzzle  is  much  less  in  evidence  than  in  most  American 
textbooks.  The  exercises  are  systematically  grouped,  appear- 
ing in  full  pages,  in  large  type,  at  frequent  intervals.  They 
are  not  all  intended  for  one  class,  but  are  so  numerous  as  to 
allow  the  teacher  to  make  selections  from  year  to  year. 

6.  The  introduction  has  been  made  as  concrete  as  is  reason- 
able. Definitions  have  been  postponed  until  they  are  actually 
needed,  only  well-recognized  terms  have  been  employed,  the 
pupil  is  initiated  at  once  into  the  practical  use  of  the  instru- 
ments, some  of  the  reasons  for  studying  geometry  are  early 
shown  in  an  interesting  way,  and  correlation  is  made  with 
the  simple  algebra  already  studied. 

The  authors  are  indebted  to  many  friends  of  the  Wentworth 
Geometry  for  assistance  and  encouragement  in  the  labor  of  pre- 
paring this  edition,  and  they  will  welcome  any  further  sugges- 
tions for  improvement  from  any  of  their  readers. 

GEORGE  WENTWORTH 
DAVID  EUGENE  SMITH 


CONTENTS 


Page 

INTRODUCTION 1 

BOOK  I.    RECTILINEAR  FIGURES 25 

Triangles 26 

Parallel  Lines 46 

Quadrilaterals       .........  59 

Polygons 68 

Loci 73 

BOOK  II.    THE  CIRCLE 93 

Theorems 94 

Problems 126 

BOOK  III.    PROPORTION.    SIMILAR  POLYGONS  .         .         .151 

Theorems 152 

Problems 182 

BOOK  IV.    AREAS  OF  POLYGONS 191 

Theorems 192 

Problems 214 

BOOK  V.    REGULAR  POLYGONS  AND  CIRCLES           .         .  227 

Theorems 228 

Problems 242 

APPENDIX 261 

Symmetry 261 

Maxima  and  Minima        ........  265 

Recreations 273 

History  of  Geometry    ........  277 

INDEX 283 

V 


SYMBOLS  AND  ABBREVIATIONS 


=   equals,  equal,  equal  to, 

Adj. 

adjacent. 

is  equal  to,  or 

Alt. 

alternate. 

is  equivalent  to. 

Ax. 

axiom. 

>   is  greater  than. 

Const. 

construction. 

<  is  less  than. 

Cor. 

corollary. 

II  parallel. 

Def. 

definition. 

_L  perpendicular. 

Ex. 

exercise. 

Z  angle. 

Ext. 

exterior. 

A  triangle. 

Fig. 

figure. 

O  parallelogram. 

Hyp. 

hypothesis. 

□  rectangle. 

Iden. 

identity. 

O  circle. 

Int. 

interior. 

st.  straight. 

Post. 

postulate. 

rt.  right. 

Prob. 

problem. 

•.•  since. 

Prop. 

proposition. 

.'.  therefore. 

Sup. 

supplementary. 

These  symbols  take  the  plural  form  when  necessary,  as  in  the  case  of 


The  symbols  +,  — ,  x,  -^  are  used  as  in  algebra. 

There  is  no  generally  accepted  symbol  for  "is  congruent  to,"  and  the 
words  are  used  in  this  book.  Some  teachers  use  =  or  =,  and  some  use 
=  ,  but  the  sign  of  equality  is  more  commonly  employed,  the  context 
telling  whether  equality,  equivalence,  or  congruence  is  to  be  understood. 

Q.  E.  D.  is  an  abbreviation  that  has  long  been  used  in  geometry  for 
the  Latin  words  quod  erat  demonstrandum,  "which  was  to  be  proved." 

Q.  E.  F.  stands  for  quod  erat  faciendum,  "which  was  to  be  done." 


PLANE  GEOMETRY 


INTRODUCTION 

1.  The  Nature  of  Arithmetic.  In  arithmetic  we  study  compu- 
tation, the  working  with  numbers.  We  may  have  a  formula 
expressed  in  algebraic  symbols,  such  as  a  =  hh,  where  a  may 
stand  for  the  area  of  a  rectangle,  and  h  and  h  respectively  for 
the  number  of  units  of  length  in  the  base  and  height ;  but  the 
actual  computation  involved  in  applying  such  formula  to  a 
particular  case  is  part  of  arithmetic. 

2.  The  Nature  of  Algebra.  In  algebra  we  generalize  the 
arithmetic,  and  instead  of  saying  that  the  area  of  a  rectangle 
with  base  4  in.  and  height  2  in.  is  4  x  2  sq.  in.,  we  express  a 
general  law  by  saying  that  a  =  hh.  In  arithmetic  we  may  have 
an  equality,  like  2  x  16  +17=  49,  but  in  algebra  we  make  much 
use  of  equations,  like  2  cc  -f- 17  =  49.  Algebra,  therefore,  is  a 
generalized  arithmetic. 

3.  The  Nature  of  Geometry.  We  are  now  about  to  begin  another 
branch  of  mathematics,  one  not  chiefly  relating  to  numbers 
although  it  uses  numbers,  and  not  primarily  devoted  to  equa- 
tions although  using  them,  but  one  that  is  concerned  principally 
with  the  study  of  forms,  such  as  triangles,  parallelograms,  and 
circles.  Many  facts  that  are  stated  in  arithmetic  and  algebra 
are  proved  in  geometry.  For  example,  in  geometry  it  is  proved 
that  the  square  on  the  hypotenuse  of  a  right  triangle  equals 
the  sum  of  the  squares  on  the  other  two  sides,  and  that  the 
circumference  of  a  circle  equals  3.1416  times  the  diameter, 

1 


PLANE  GEOMETKY 


4.  Solid.  The  block  here  represented  is  called  a  solid;  it 
is  a  limited  portion  of  space  filled  with  matter.  In  geometry, 
however,  we  have  nothing  to  do  with  the  matter  of  which  a 


body  is  composed;  we  study  simply  its  sha^je  and  size,  as  in 
the  second  figure. 

That  is,  a  physical  solid  can  be  touched  and  handled ;  a  geometric 
solid  is  the  space  that  a  physical  solid  is  conceived  to  occupy.  For 
example,  a  stick  is  a  physical  solid  ;  but  if  we  put  it  into  wet  plaster,  and 
then  remove  it,  the  hole  that  is  left  may  be  thought  of  as  a  geometric 
solid  although  it  is  filled  with  air. 

5.  Geometric  Solid.  A  limited  portion  of  space  is  called  a 
geometric  solid. 

6.  Dimensions.  The  block  represented  in  §  4  extends  in  three 
principal  directions : 

(1)  From  left  to  right,  that  is,  from  A  to  D] 

(2)  From  back  to  front,  that  is,  from  A  to  B] 

(3)  From  top  to  bottom,  that  is,  from  A  to  E. 

These  extensions  are  called  the  dimensions  of  the  block,  and 
are  named  in  the  order  given,  length,  breadth  (or  width),  and 
thickness  (height,  altitude,  or  depth).  Similarly,  we  may  say 
that  every  solid  has  three  dimensions. 

Very  often  a  solid  is  of  such  shape  that  we  cannot  point  out  the  length, 
or  distinguish  it  from  the  breadth  or  thickness,  as  an  irregular  block  of 
coal.  In  the  case  of  a  round  ball,  where  the  length,  breadth,  and  thick- 
ness are  all  the  same  in  extent,  it  is  impossible  to  distinguish  one  dimen- 
sion from  the  others. 


INTEODUCTIO>T  3 

7.  Surface.  The  block  shown  in  §  4  has  six  flat  faces,  each 
of  which  is  called  a  surface.  If  the  faces  are  made  smooth  by 
polishing,  so  that  when  a  straight  edge  is  applied  to  any  one 
of  them  the  straight  edge  in  every  part  will  touch  the  surface, 
each  face  is  called  2^  plane  surface,  or  2^  plane. 

These  surfaces  are  simply  the  boundaries  of  the  solid.  They  have  no 
thickness,  even  as  a  colored  light  shining  upon  a  piece  of  paper  does  not 
make  the  paper  thicker.  A  board  may  be  planed  thinner  and  thinner, 
and  then  sandpapered  still  thinner,  thus  coming  nearer  and  nearer  to 
representing  what  we  think  of  as  a  geometric  plane,  but  it  is  always  a 
solid  bounded  by  surfaces. 

That  which  has  length  and  breadth  without  thickness  is 
called  a  surface. 

8.  Line.  In  the  solid  shown  in  §  4  we  see  that  two  adja- 
cent surfaces  intersect  in  a  line.  A  line  is  therefore  simply 
the  boundary  of  a  surface,  and  has  neither  breadth  nor  thickness. 

That  which  has  length  without  breadth  or  thickness  is  called 
a  line. 

A  telegraph  wire,  for  example,  is  not  a  line.  It  is  a  solid.  Even  a 
pencil  mark  has  width  and  a  very  little  thickness,  so  that  it  is  also  a  solid. 
But  if  we  think  of  a  wire  as  drawn  out  so  that  it  becomes  finer  and  finer, 
it  comes  nearer  and  nearer  to  representing  what  we  think  of  and  speak 
of  as  a  geometric  line. 

9.  Magnitudes.  Solids,  surfaces,  and  lines  are  called  tnag- 
nitudes. 

10.  Point.  In  the  solid  shown  in  §  4  we  see  that  when  two 
lines  meet  they  meet  in  a  point.  A  point  is  therefore  simply 
the  boundary  of  a  line,  and  has  no  length,  no  breadth,  and 
no  thickness. 

That  which  has  only  position,  without  length,  breadth,  or 
thickness,  is  called  di:  point. 

We  may  think  of  the  extremity  of  a  line  as  a  point.  We  may  also 
think  of  the  intersection  of  two  lines  as  a  point,  and  of  the  intersection 
of  two  surfaces  as  a  line. 


4  PLANE  GEOMETRY 

11.  Representing  Points  and  Geometric  Magnitudes.  Although 
we  only  imagine  such  geometric  magnitudes  as  lines  or  planes, 
we  may  represent  them  by  pictures. 

Thus  we  represent  a  point  by  a  fine  dot,  and  I  \        7 

name  it  by  a  letter,  as  P  in  this  figure.  /  / 

We  represent  a  hne  by  a  fine  mark,  and  name       L ' 

it  by  letters  placed  at  the  ends,  as  ^5. 

We  represent  a  surface  by  its  boundary  lines,  and  name  it  by  letters 
placed  at  the  corners  or  in  some  other  convenient  way,  as  A  BCD. 

We  represent  a  solid  by  the  boundary  faces  or  by  the  lines  bounding 
the  faces,  as  in  §  4. 

12.  Generation  of  Geometric  Magnitudes.    We  may  think  of 

(1)  A  line  as  generated  by  a  moving  point ; 

(2)  A  surface  as  generated  by  a  moving  line ; 

(3)  A  solid  as  generated  by  a  moving  surface. 

For  example,  as  shown  in  the  figure  let  the  surface  A  BCD  move  to 
the  position  WXYZ.    Then 

(1)  A  generates  the  line  AW; 

(2)  AB  generates  the  surface  A  WXB  ;      D 

(3)  ABCD  generates  the  solid  AY. 


C         ., Y 


A 


— >— 
B 


71 


->■ 


Of  course  a  point  will  not  generate  a  line         /        '  / 

by  simply  turning  over,  for  this  is  not  mo-      ^  ^  ^ 

tion  for  a  point ;  nor  will  a  line  generate  a 

surface  by  simply  sliding  along  itself  ;  nor  will  a  surface  generate  a  solid 
by  simply  sliding  upon  itself. 

13.  Geometric  Figure.  A  point,  a  line,  a  surface,  a  solid,  or 
any  combination  of  these,  is  called  a  geometric  figure. 

A  geometric  figure  is  generally  called  simply  a  figure. 

14.  Geometry.  The  science  of  geometric  figures  is  called 
geometry. 

Plane  geometry  treats  of  figures  that  lie  wholly  in  the  same 
plane,  that  is,  of  plane  figures. 

Solid  geometry  treats  of  figures  that  do  not  lie  wholly  in 
the  same  plane. 


INTEODUCTIOK 


15.  Straight  Line.  A  line  such  that  any  part  placed  with  its 
ends  on  any  other  part  must  lie  wholly  in  the  line  is  called  a 
straight  line. 

For  example,  J.i>  is  a  straight  line,  for  if  we  take,  say,  a  half  inch  of  it, 

and  place  it  in  any  way  on  any  other  part  of  AB,     

but  so  that  its  ends  lie  in  AB^  then  the  whole  of 

the  half  inch  of  line  will  lie  in  AB.    This  is  well  shown  by  using  tracing 

paper.    The  word  line  used  alone  is  understood  to  mean  a  straight  line. 

Part  of  a  straight  line  is  called  a  segment  of  the  line.  The  term  seg- 
ment is  applied  also  to  certain  other  magnitudes. 

16.  Equality  of  Lines.  Two  straight-line  segments  that  can 
be  placed  one  upon  the  other  so  that  their  extremities  coin- 
cide are  said  to  be  equal. 

In  general,  two  geometric  magnitudes  are  equal  if  they  can  be 
made  to  coincide  throughout  their  whole  extent.  We  shall  see  later 
that  some  figures  that  coincide  are  said  to  be  congruent. 

17.  Broken  Line.  A  line  made  up  of 
two  or  more  different  straight  lines  is 
called  a  broken  line. 

For  example,  CD  is  a  broken  line. 

18.  Rectilinear  Figure.  A  plane  figure 
bounded  by  a  broken  line  is  called  a  rec- 
tilinear figure. 

For  example,  A  BCD  is  a  rectilinear  figure. 

19.  Curve  Line.  A  line  no  part  of  which 
is  straight  is  called  a  curiae  line,  or  simply 
a  curve. 

For  example,  EF  is  a  curve  line. 

20.  Curvilinear  Figure.  A  plane  figure  formed 
by  a  curve  line  is  called  a  curvilinear  figure. 

For  example,  0  is  a  curvilinear  figure  with  which 
we  are  already  familiar. 

Some  curvilinear  figures  are  surfaces  bounded  by 
curves  and  others  are  the  curves  themselves. 


•O 


6 


PLANE  GEOMETEY 


21.  Angle.  The  opening  between  two  straight  lines  drawn 
from  the  same  point  is  called  an  angle. 

Strictly  speaking,  this  is  a  plane  angle.  We  shall 
find  later  that  there  are  angles  made  by  curve  lines  and 
angles  made  by  planes. 

The  two  lines  are  called  the  sides  of  the  angle,  and    q 
the  point  of  meeting  is  called  the  vertex. 

An  angle  may  be  read  by  naming  the  letters  desig- 
nating the  sides,  the  vertex  letter  being  between  the 
others,  as  the  angle  A  OB.  An  angle  may  also  be  desig- 
nated by  the  vertex  letter,  as  the  angle  O,  or  by  a  small 
letter  within,  as  the  angle  m.  A  curve  is  often  drawn  to  show  the  par- 
ticular angle  meant,  as  in  angle  m. 

22.  Size  of  Angle.  The  size  of  an  angle  depends  upon  the 
amount  of  turning  necessary  to  bring  one  side  into  the  position 
of  the  other. 

One  angle  is  greater 
than  another  angle  when 
the  amount  of  turning  is 
greater.  Thus  in  these 
compasses  the  first  angle 


is  smaller  than  the  second,  which  is  also  smaller  than  the  third, 
length  of  the  sides  has  nothing  to  do  with  the  size  of  the  angle. 


The 


23.  Equality  of  Angles.  Two  angles  that  can  be  placed  one 
upon  the  other  so  that  their  vertices  coincide  and  the  sides  of 
one  lie  along  the  sides  of  the  other  are  said  to  ^b 
be  equal. 

For  example,  the  angles  AOB  and  A'O'B'  (read 
"J.  prime,  0  prime,  B  prime")  are  equal.  It  is  well 
to  illustrate  this  by  tracing  one  on  thin  paper  and 
placing  it  upon  the  other.  q'^c ^ 

24.  Bisector.  A  point,  a  line,  or  a  plane  that  divides  a  geo- 
metric magnitude  into  two  equal  parts  is  called  a  bisector  of 
the  magnitude. 

For  example,  M,  the  mid-point  of  the  line  AB^       A  M  B 

is  a  bisector  of  the  line. 


INTRODUCTION 


25.  Adjacent  Angles.  Two  angles  that  have  the  same  vertex 
and  a  common  side  between  them  are  called  adjacent  angles. 

For  example,  the  angles  AOB  and  BOC  are 
adjacent  angles,  and  in  §26  the  angles  AOB  and 
BOC  are  adjacent  angles. 

26.  Right  Angle.    When  one  straight  line 

meets  another  straight  line  and  makes  the 

adjacent  angles  equal,  each  angle  is  called  a 

right  angle. 

For  examplie,  angles  A  OB  and  BOCm  this  figure. 
If  CO  is  cut  off,  angle  AOB  is  still  a  right  angle.         ^  ^  ^ 

27.  Perpendicular.  A  straight  line  making  a  right  angle  with 
another  straight  line  is  said  to  be  ijeirpendicular  to  it. 

Thus  OB  is  perpendicular  to  CA^  and  CA  to  OB.  OB  is  also  called  a 
perpendicular  to  CJ.,  and  0  is  called  the  foot  of  the  perpendicular  OB. 

28.  Triangle.  A  portion  of  a  plane  bounded  by  three  straight 
lines  is  called  a  triangle.  C 

The  lines  AB,  BC,  and  CA  are  called  the  sides 
of  the  triangle  ABC^  and  the  sides  taken  together 
form  the  perimeter.  The  points  A,  B,  and  C  are 
the  vertices  of  the  triangle,  and  the  angles  A,  B,  and  C  are  the  angles  of 
the  triangle.  The  side  AB  upon  which  the  triangle  is  supposed  to  rest 
is  the  base  of  the  triangle.    Similarly  for  other  plane  figures. 

29.  Circle.  A  closed  curve  lying  in  a  plane,  and  such  that 
all  of  its  points  are  equally  distant  from  a  fixed  point  in  the 
plane,  is  called  a  circle. 

The  length  of  the  circle  is  called  the  circumference. 
The  point  from  which  all  points  on  the  circle  are 
equally  distant  is  the  center.  Any  portion  of  a  circle 
is  an  arc.  A  straight  line  from  the  center  to  the  circle 
is  a  radius.  A  straight  line  through  the  center,  termi- 
nated at  each  end  by  the  circle,  is  a  diameter. 

Formerly  in  elementary  geometry  circle  was  taken  to  mean  the  space 
inclosed,  and  the  bounding  line  was  called  the  circumference.  Modern 
usage  has  conformed  to  the  definition  used  in  higher  mathematics. 


8 


PLANE  GEOMETRY 


30.  Instruments  of  Geometry.  In  geometry  only  two  instru- 
ments are  necessary  besides  pencil  and  paper.  These  are  a 
straight  edge,  or  ruler,  and  a  pair  of  compasses. 

It  is  evident  that  all  radii  of  the  same  circle  are  equal. 


In  the  absence  of  compasses,  and  particularly  for  blackboard  work,  a 
loop  made  of  string  may  be  used.  For  the  accurate  transfer  of  lengths, 
however,  compasses  are  desirable. 

31.  Exercises  in  using  Instruments.  The  following  simple 
exercises  are  designed  to  accustom  the  pupil  to  the  use  of 
instruments.  No  proofs  are  attempted,  these  coming  later  in 
the  course. 

This  section  may  be  omitted  if  desired,  without  affecting  the  course. 


*.CL 


EXERCISE  1 

1.  From  a  given  point  on  a  given  straight  line  required  to 
draw  a  perpendicular  to  the  line. 

Let  AB  he  the  given  line  and  P  be  the 
given  point. 

It  is  required  to  draw  from  P  a  line  per- 
pendicular to  AB. 

With  P  as  a  center  and  any  convenient 
radius  draw  arcs  cutting  AB  2it  X  and  Y. 

With  JT  as  a  center  and  XY  as  a  radius  draw  a  circle,  and  with  Y 
as  a  center  and  the  same  radius  draw  another  circle,  and  call  one  inter- 
section of  the  circles  C. 

With  a  straight  edge  draw  a  line  from  P  to  C,  and  this  will  be  the 
perpendicular  required. 


\X 


Yl 


INTRODUCTION 


9 


A  X^' 


2.  From  a  given  point  outside  a  given  straight  line  required 
to  let  fall  a  perpendicular  to  the  line.  p 

Let  AB  he  the  given  straight  line  and  P  be  the 
given  point. 

It  is  required  to  draw  from  P  a  line  perpen- 
dicular to  AB. 

With  P  as  a  center  and  any  convenient  radius 
draw  an  arc  cutting  AB  at  X  and  Y. 

With  X  as  a  center  and  any  convenient  radius 
draw  a  circle,  and  with  F  as  a  center  and  the  same 
radius  draw  another  circle,  and  call  one  intersection  of  the  circles  C. 

With  a  straight  edge  draw  a  straight  line  from  P  to  C,  and  this  will 
be  the  perpendicular  required. 

It  is  interesting  to  test  the  results  in  Exs.  1  and  2,  by  cutting  the  paper 
and  fitting  the  angles  together. 


-S^' 


I 


3.  Required  to  draw  a  triangle  having  two  sides  each  equal 
to  a  given  line. 

Let  I  be  the  given  line. 

It  is  required  to  draw  a  triangle  having  two  sides 
each  equal  to  I. 

With  any  center,  as  C,  and  a  radius  equal  to  I 
draw  an  arc. 

Join  any  two  points  on  the  arc,  as  A  and  J5, 
with  each  other  and  with  C  by  straight  lines. 

Then  ABC  is  the  triangle  required. 


^'-~ 


4.  Required  to  draw  a  triangle  having  its  three  sides  each 
equal  to  a  given  line.  ^   ^  ^ 

Let  AB  he  the  given  line. 

It  is  required  to  draw  a  triangle  having  its  three 
sides  each  equal  to  AB. 

With  A  as  a  center  and  ^Z?  as  a  radius  draw  a 
circle,  and  with  B  as  a  center  and  the  same  radius 
draw  another  circle. 

Join  either  intersection  of  the  circles  with  A  and  B  by  straight  lines. 

Then  ABC  is  the  triangle  required. 

In  such  cases  draw  the  arcs  only  long  enough  to  show  the  point  of 
intersection. 


10 


PLANE  GEOMETPvY 


^ 


^s.O.'-- 


5.  Eequired  to  draw  a  triangle  having  its  sides  equal  respec- 
tively to  three  given  lines. 

Let  the  three  lines  be  I,  m,  and  n. 

What  is  now  required  ? 

Upon  any  line  mark  off  with  the  com- 
passes a  line-segment  AB  equal  to  I. 

With  ^  as  a  center  and  m  as  a  radius 
draw  a  circle ;  with  2^  as  a  center  and 
n  as  a  radius  draw  a  circle. 

Drawee  and  50.  ^ 

Then  ABC  is  the  required  triangle.  n 

6.  From  a  given  point  on  a  given  line  required  to  draw  a 
line  making  an  angle  equal  to  a  given  angle. 


X 


^^-- 


o 


\c 


IM 


Let  P  be  the  given  point  on  the  given  line  PQ,  and  let  angle  AOB  be 
the  given  angle. 

What  is  now  required  ? 

With  0  as  a  center  and  any  radius  draw  an  arc  cutting  ^  O  at  C  and 
BO  at  D. 

With  P  as  a  center  and  OC  as  a  radius  draw  an  arc  cutting  PQ  at  M. 

With  Jlf  as  a  center  and  the  straight  line  joining  C  and  D  as  a  radius 
draw  an  arc  cutting  the  arc  just  drawn  at  N,  and  draw  PN. 

Then  angle  MPN  is  the  required  angle. 


7.  Eequired  to  bisect  a  given  straight  line. 

Let  AB  be  the  given  line. 

It  is  required  to  bisect  it. 

With  J.  as  a  center  and  ^B  as  a  radius  draw  a  circle, 
and  with  B  as  a  center  and  the  same  radius  draw  a  circle. 

Call  the  two  intersections  of  the  circles  X  and  Y. 

Draw  the  straight  line  XY. 

Then  XY  bisects  the  line  AB  a^t  the  point  of  inter- 
section M. 


^C 


M 


-r 


INTKODUCTION 


11 


8.  Required  to  bisect  a  given  angle. 

Let  A  OB  be  the  given  angle. 

It  is  required  to  bisect  it. 

With  O  as  a  center  and  any  convenient  radius 
draw  an  arc  cutting  OA  at  X  and  OB  at  Y. 

With  X  as  a  center  and  a  line  joining  X  and 
F  as  a  radius  draw  a  circle,  and  with  F  as  a 
center  and  the  same  radius  draw  a  circle,  and  call  one  point  of  inter- 
section of  the  circles  P. 

Draw  the  straight  line  OP. 

Then  OP  is  the  required  bisector. 

9.  By  the  use  of  compasses  and  ruler  draw  the  following 
figures :  ^^    ^ 


The  dotted  lines  show  how  to  fix  the  points  needed  in  drawing  the 
figure,  and  they  may  be  erased  after  the  figure  is  completed.  In  general, 
in  geometry,  auxiliary  lines  (those  needed  only  as  aids)  are  indicated  by 
dotted  lines. 

10.  By  the  use  of  compasses  and  ruler  draw  the  following 
figures : 


It  is  apparent  from  the  figures  in  Exs.  9  and  10  that  the  radius  of 
the  circle  may  be  used  in  describing  arcs  that  shall  divide  the  circle  into 
six  equal  parts. 


12 


PLANE  GEOMETRY 


11.  By  the  use  of  compasses  and  ruler  draw  the  following 
figures : 


12.  By  the  use  of  compasses  and  ruler  draw  the  following 
figures : 


13.  By  the  use  of  compasses  and  ruler  draw  the  following 
figures : 


In  such  figures  artistic  patterns  may  be  made  by  coloring  various 
portions  of  the  drawings.  In  this  way  designs  are  made  for  stained-glass 
windows,  for  oilcloth,  for  colored  tiles,  and  for  other  decorations. 

14.  Draw  a  triangle  of  which  each  side  is  li  in. 

15.  Draw  two  lines  bisecting  each  other  at  right  angles. 


INTRODUCTION  13 

16.  Bisect  each  of  the  four  right  angles  formed  by  two  lines 
bisecting  each  other  at  right  angles. 

17.  Draw  a  line  1^-  in.  long  and  divide  it  into  eighths  of  an 
inch,  using  the  ruler.  Then  with  the  compasses  draw  this 
figure. 

It  is  easily  shown,  when  we  come  to 
the  measurement  of  the  circle,  that  these 
two  curve  lines  divide  the  space  inclosed 
by  the  circle  into  parts  that  are  exactly 
equal  to  one  another. 

By  continuing  each  semicircle  to 
make  a  complete  circle  another  inter- 
esting figure  is  formed.  Other  similar 
designs  are  easily  invented,  and  students 
should  be  encouraged  to  make  such 
original  designs. 

18.  In  planning  a  Gothic  window  this  drawing  is  needed. 
The  arc  BC  is  drawn  with  ^  as  a  center  q 

and  vlJ5  as  a  radius.  The  small  arches 
are  described  with  A,  Z),  and  B  as  centers 
and  ^i)  as  a  radius.  The  center  P  is  found 
by  taking  A  and  B  as  centers  and  AE  sls 
a  l-adius.  How  may  the  points  D,  E,  and 
F  be  found  ?    Draw  the  figure.  j_     p      ^     E      B 

19.  Draw  a  triangle  of  which  each  side  is  1  in.  Bisect  each 
side,  and  with  the  points  of  bisection  as  centers  and  with  radii 
^  in.  long  draw  three  circles. 

20.  A  baseball  diamond  is  a  square  90  ft.  on  a  side.  Draw 
the  plan,  using  a  scale  of  ^^  in.  to  a  foot.  Locate  the  pitcher 
60  ft.  from  the  home  plate. 

21.  A  man  travels  from  A  directly  east  1  mi.  to  B.  He  then 
turns  and  travels  directly  north  1|  mi.  to  C.  Draw  the  plan 
and  find  by  measurement  the  distance  ^  C  to  the  nearest  quarter 
of  a  mile.    Use  a  scale  of  ^  in.  to  a  mile. 


14  PLANE  GEOMETRY 

22.  A  double  tennis  court  is  78  ft.  long  and  36  ft.  wide.  The 
net  is  placed  39  ft.  from  each  end  and  the  service  lines  18  ft. 
from  each  end.  Draw  the  plan,  using  a  scale  of  y^^  in.  to  a  foot, 
making  the  right  angles  as  shown  in  Ex.  1.  The  accuracy  of 
the  construction  may  be  tested  by  measuring  the  diagonals, 
which  should  be  equal. 

23.  At  the  entrance  to  New  York  harbor  is  a  gun  having 
a  range  of  12  mi.  Draw  a  line  inclosing  the  range  of  fire, 
using  a  scale  of  ^^  in.  to  a  mile. 

24.  Two  forts  are  placed  on  opposite  sides  of  a  harbor 
entrance,  13  mi.  apart.  Each  has  a  gun  having  a  range  of 
10  mi.  Draw  a  -plan  showing  the  area  exposed  to  the  fire  of 
both  guns,  using  a  scale  of  ^\  in.  to  a  mile. 

25.  Two  forts,  A  and  B,  are  placed  on  opposite  sides  of  a 
harbor  entrance,  16  mi.  apart.  On  an  island  in  the  harbor,  12 
mi.  from  A  and  11  mi.  from  B,  is  a  fort  C.  The  fort  A  has  a 
gun  with  a  range  of  12  mi.,  fort  B  one  with  a  range  of  11  mi., 
and  fort  C  one  with  a  range  of  10  mi.  Draw  a  plan  of  the 
entrance  to  the  harbor,  showing  the  area  exposed  to  the  fire 
of  each  gun. 

26.  A  horse,  tied  by  a  rope  25  ft.  long  at  the  corner  of  a  lot 
50  ft.  square,  grazes  over  as  much  of  the  lot  as  possible.  The 
next  day  he  is  tied  at  the  next  corner,  the  third  day  at  the 
third  corner,  and  the  fourth  day  at  the  fourth  corner.  Draw 
a  plan  showing  the  area  over  which  he  has  grazed  during  the 
four  days,  using  a  scale  of  ^  in.  to  5  ft. 

27.  A  gardener  laid  out  a  flower  bed  on  the  following  plan  : 
He  made  a  triangle  ABC,  16  ft.  on  a  side,  and  then  bisected 
two  of  the  angles.  From  the  point  of  intersection  of  the  bi- 
sectors, P,  he  drew  perpendiculars  to  the  three  sides  of  the 
triangle,  PX,  PY,  and  PZ.  Then  he  drew  a  circle  with  P  as  a 
center  and  PX  as  a  radius,  and  found  that  it  just  fitted  in  the 
triangle.    Draw  the  plan,  using  a  scale  of  i  in.  to  a  foot. 


IKTEODUCTIOK 


15 


32.  Necessity  for  Proof.  Although  pUrt  of  geometry  consists 
in  drawing  figures,  this  is  not  the  most  important  part.  It  is 
essential  to  prove  that  the  figures  are  what  we  claim  them  to  be. 
The  danger  of  trusting  to  appearances  is  seen  in  Exercise  2. 


EXERCISE  2 

1.  Estimate  which  is  the  longer  line,  AB  or  XF,  and  how 
much  longer.    Then  test  your  estimate  by     ^> <^B 


measuring  with  the  compasses  or  with  a 
piece  of  paper  carefully  marked. 

2.  Estimate  which  is  the  longer  line,  AB  ot 
CDj  and  how  much  longer.  Then  test  your 
estimate  by  measuring  as  in  Ex.  1. 

3-  Look  at  this  figure  and  state  whether 
AB  and  CD  are  both  straight  lines.  If  one 
is  not  straight,  which  one  is 
it?  Test  your  answer  by  us- 
ing a  ruler  or  the  folded  edge 
of  a  piece  of  paper. 

4.  Look  at  this  figure  and 
state  whether  AB  and  CD  are 
the  same  distance  apart  at  .4  and  C  as 
at  B  and  D.    Then  test  your  answer  as 
in  Ex.  1. 

5 .  Look  at  this  figure  and  state  whether 
AB  will,  if  prolonged,  lie  on  CD.  Also 
state  whether  WX  will,  if  prolonged,  lie 
on  YZ.    Then  test  your  answer 


<- 


^mmm^}^ 


by  laying  a  ruler  along  the  lines.    ^^ 

6.  Look  at  this  figure  and  state  which 
of  the  three  lower  lines  is  ^-B  prolonged. 
Then  test  your  answer  by  laying  a  ruler 
along  AB, 


i 


16  PLANE  GEOMETRY 

33.  Straight  Angle.  When  the  sides  of  an  angle  extend  in 
opposite  directions,  so  as  to  be  in  the  same  straight  line,  the 
angle  is  called  a  straight  angle.  ^-^ 

For  example,  the  angle  A  OB,  as  shown  in  this  ^ 

figure,  is  a  straight  angle.    The  angle  BOA,  below  the  line,  is  also  a 
straight  angle. 

34.  Right  Angle  and  Straight  Angle.  It  follows  from  the 
definition  of  right  angle  (§  26)  that  a  right  angle  is  half  of  a 
straight  angle. 

In  like  manner,  it  follows  that  a  straight  angle  equals  twice 
a  right  angle. 

35.  Acute  Angle.  An  angle  less  than  a  right  angle  is  called 
an  aciite  angle. 

For  example,  the  angle  m,  as  shown  in  this  figure,  is 
an  acute  angle. 

36.  Obtuse  Angle.  An  angle  greater  than  a  right  angle  and 
less  than  a  straight  angle  is  called  an  obtuse  „ 

angle. 

For  example,  the  angle  AOB,  as  shown  in  this  .    , 

figure,  is  an  obtuse  angle.  ''^9^y 

37.  Reflex  Angle.  An  angle  greater  than  a  straight  angle 
and  less  than  two  straight  angles  is  called  a  reflex  angle. 

For  example,  the  angle  BOA,  marked  with  a  dotted  curve  line  in  the 
figure  in  §  36,  is  a  reflex  angle. 

When  we  speak  of  an  angle  formed  by  two  given  lines  drawn  from  a 
point  we  mean  the  smaller  angle  unless  the  contrary  is  stated. 

38.  Oblique  Angles.  Acute  angles  and  obtuse  angles  are  called 
oblique  angles. 

The  sides  of  oblique  angles  are  said  to  be  oblique  to  each 
other,  and  are  called  oblique  lines. 

Evidently  if  we  bisect  a  straight  angle,  we  form  two  right  angles ;  if 
we  bisect  a  right  angle  or  an  obtuse  angle,  we  form  two  acute  angles  ; 
if  we  bisect  a  reflex  angle,  we  form  two  obtuse  angles. 


INTEODUCTION 


17 


39.  Generation  of  Angles.  Suppose  the  line  r  to  revolve  from 
the  position  OA  about  the  point  O  as  a  vertex  to  the  posi- 
tion OB.  Then  r  describes  or  generates 
the  acute  angle  A  OB,  and,  as  we  have  seen 
(§  22)  the  size  of  the  angle  depends  upon  the 
amount  of  rotation,  the  angle  being  greater 
as  the  amount  of  turning  is  greater. 

If  r  rotates  still  further,  to  the  position  OC,  it 
has  then  generated  the  right  angle  AOC  and  is 
perpendicular  to  OA. 

If  r  rotates  still  further,  to  the  position  OD,  it  has  then  generated  the 
obtuse  angle  A  OB. 

If  r  rotates  to  the  position  OE^  it  has  then  generated  the  straight 
angle  AOE. 

If  r  rotates  to  the  position  OF,  it  has  then  generated  the  reflex 
angle  A  OF. 

If  r  rotates  still  further,  past  06?  to  the  position  OA  again,  it  has 
made  a  complete  revolution  and  has  generated  two  straight  angles  or 
four  right  angles. 


40.  Sums  and  Differences  of  Magnitudes.    If  the  straight  line 

AP  has  been  generated  by  a  point  P     , i___+__>P 

moving  from  A   to  P,  the   segments   ^ 


t 


B         CD 

AB,  BC,  CD,  and  so  on,  having  been  generated  in  succession, 
then  we  call  AC  the  stem  oi  AB  and  BC.    That  is, 

AC  =  AB-\- BC,  whence  AC  —  BC  =  AB. 

If  the  angle  A  OD  has  been  generated  by  the  -  (^ 
line  OA  revolving  about  0  as  a  vertex  from 
the  position  OA,  the  angles  AOB,  BOC,  and 
COD  having  been  generated  in  succession, 
then  we  call  angle  AOC  the  sum  of  angles 
AOB  and  BOC.    That  is,  considering  angles, 

.4  OC  =  ^  0^  +  BOC,  whence  AOC  -  BOC  =  A  OB. 

In  the  same  way  that  we  may  have  the  sum  or  the  difference  of  lines 
or  of  angles  we  may  have  the  sum  or  the  difference  of  surfaces  or  of  solids. 


18  PLANE  GEOMETRY 

41.  Perigon.  The  whole  angular  space  in  a  plane  about  a 
point  is  called  a  perigon. 

It  therefore  follows  that  a  perigon  equals  the  sum  of  two  straight 
angles  or  the  sum  of  four  right  angles. 

42.  Complements,  Supplements,  and  Conjugates,     If  the  sum 

of  two  angles  is  a  right  angle,  each  angle  is  called  the  comple- 
ment of  the  other. 

If  the  sum  of  two  angles  is  a  straight  angle,  each  angle  is 
called  the  supplement  of  the  other. 

If  the  sum  of  two  angles  is  a  perigon,  each 
angle  is  called  the  conjugate  of  the  other. 

Thus,  with  respect  to  angle  AOB, 
the  complement  is  angle  BOC, 
the  supplement  is  angle  J50D, 
the  conjugate  is  angle  BOA  (reflex). 

43.  Properties  of  Supplementary  Angles.  It  is  sufficiently  evi- 
dent to  be  taken  without  proof  that 

1.  The  two  adjacent  angles  which  one  straight  line  Tuakes  with 
another  are  together  equal  to  a  straight  angle. 

2.  If  the  sum  of  two  adjacent  angles  is  a  straight  angle,  their 
exterior  sides  are  in  the  sam,e  straight  line. 

44.  Angle  Measure.  Angles  are  measured  by  taking  as  a  unit 
•j^-Q  of  a  perigon.    This  unit  is  called  a  degree. 

The  degree  is  divided  into  60  equal  parts,  called  minutes,  and 
the  minute  into  60  equal  parts,  called  seconds. 

We  write  5°  13'  12"  for  5  degrees  13  minutes  12  seconds. 

It  is  evident  that  a  right  angle  equals  90°,  a  straight  angle  equals  180°, 
and  a  perigon  equals  360°. 

45.  Vertical  Angles.  When  two  angles  have  the  same  vertex, 
and  the  sides  of  the  one  are  prolongations  of 

the  sides  of  the  other,  those  angles  are  called        \ 

vertical  angles.  ^\" 

In  the  figure  the  angles  x  and  z  are  vertical  angles,  \ 

as  are  also  the  angles  w  and  y.  \ 


INTRODUCTION  19 

EXERCISE  3 

1.  Find  the  complement  of  72°;  of  65^30';  of  22°  20' 15". 

2.  What  is  the  supplement  of  45°  ?   of  120°  ?  of  145°  5'  ? 
of  22°  20'  15"  ? 

3.  What  is  the  conjugate  of  240°  ?  of  280°  ?  of  312°  10'  40"  ? 

4.  The  complement  of  a  certain  angle  x  is  2x. 
How  many  degrees  are  there  in  x  ? 


yx 


5.  The  complement  of  a  certain  angle  ic  is  3ic.    How  many 
degrees  are  there  in  a?  ? 

6.  What  is  the  angle  of  which  the  complement  is  four  times 
the  angle  itself  ? 

7.  The  supplement  of  a  certain  angle  a;  is  5  ic.                  ^^ 
How  many  degrees  are  there  in  ic  ?  ^xy^ 

8.  The  supplement  of  a  certain  angle  x  is  14  x.    How  many 
degrees  are  there  in  cc  ? 

9.  What  is  the  angle  of  which  the  supplement  equals  half 
of  the  angle  itself  ? 

10.  How  many  degrees  in  an  ahgle  that  equals  its  own  com- 
plement ?  in  one  that  equals  its  own  supplement  ? 

11.  The  conjugate  of  a  certain  angle  ic  is  fee.        A  ^    A 

How  many  degrees  are  there  in  cc  ?  y^ 

12.  The  conjugate  of  a  certain  angle  x  is  \x.    How  many 
degrees  are  there  in  cc  ? 

13.  How  many  degrees  in  an  angle  that  equals  a  third  of  its 
own  conjugate?  in  one  that  equals  its  own  conjugate? 

14.  Find  two  angles,  x  and  y,  such  that  their  sum  is  90°  and 
their  (difference  is  10°. 

15.  Find  two  complementary  angles  such  that  their  differ- 
ence is  30°. 

16.  Find  two  supplementary  angles   such  that   one  is  20° 
greater  than  the  other. 


20  PLANE  GEOMETRY 

17.  The  angles  x  and  y  are  conjugate  angles,  and  their  differ- 
ence is  a  straight  angle.    How  many  degrees  are  there  in  each  ? 

18.  The  angles  x  and  y  are  conjugate  angles,  and  their  differ- 
ence is  zero.    How  many  degrees  are  there  in  each  ? 

19.  Of  two  complementary  angles  one  is  four  fifths  of  the 
other.    How  many  degrees  are  there  in  each  ? 

20.  Of  two  supplementary  angles  one  is  five  times  the  other. 
How  many  degrees  are  there  in  each  ? 

21.  How  many  degrees  are  there  in  the  smaller  angle  formed 
by  the  hands  of  a  clock  at  5  o'clock  ? 

22.  How  many  degrees  are  there  in  the  smaller  angle  formed 
by  the  hands  of  a  clock  at  10  o'clock  ?  b 

23.  In  this  figure,  if  angle  A  OB  is  38°,  how  y^ 
many  degrees  in  angle  BOC  ?    How  many  in            Xo 
angle  COD?    How  many  in  angle  DO  A  ?                j{ 

24.  In  the  same  figure,  if  angle  AOB  is  equal  to  a  third  of 
angle  BOC,  how  many  degrees  in  each  of  the  four  angles  ? 

25.  In  the  angles  of  this  figure,  \iw  =  2x,  how 
many  degrees  in  each  ?   How  many  degrees  iny?         \ 

How  many  degrees  in  ^  ?  A^ 

26.  Find    the    angle    whose    complement    de-  \ 
creased  by  30°  equals  the  angle  itself. 

27.  Find  the  angle  whose  complement  divided  by  2  equals 
the  angle  itself. 

28.  Draw  a  figure  to  show  that  if  two  adjacent  angles  have 
their  exterior  sides  in  the  same  straight  line,  their  sum  is  a 
straight  angle. 

29.  Draw  a  figure  to  show  that  the  sum  of  all  the  angles 
on  the  same  side  of  a  straight  line,  at  a  given  point,  is  equal 
to  two  right  angles. 

30.  Draw  a  figure  to  show  that  the  complements  of  equal 
angles  are  equal. 


INTRODUCTION  21 

46.  Axiom.  A  general  statement  admitted  without  proof  to 
be  true  is  called  an  axiom. 

For  example,  it  is  stated  in  algebra  that  "if  equals  are  added  to 
equals  the  sums  are  equal."  This  is  so  simple  that  it  is  generally  accepted 
without  proof.    It  is  therefore  an  axiom. 

47.  Postulate.  In  geometry  a  geometric  statement  admitted 
without  proof  to  be  true  is  called  a  postulate. 

For  example,  it  is  so  evident  that  all  straight  angles  are  equal,  that 
this  statement  is  a  postulate.  It  is  also  evident  that  a  straight  line  may 
be  drawn  and  that  a  circle  may  be  described,  and  these  statements  are 
therefore  postulates  of  geometry. 

Axioms  are  therefore  general  mathematical  assumptions,  while  geo- 
metric postulates  are  the  assumptions  peculiar  to  geometry.  Postulates 
and  axioms  are  the  assumptions  upon  which  the  whole  science  of  mathe- 
matics rests. 

48.  Theorem.    A  statement  to  be  proved  is  called  a  theorem. 

For  example,  it  is  stated  in  arithmetic  that  the  square  on  the  hypote- 
nuse of  a  right  triangle  equals  the  sum  of  the  squares  on  the  other  two 
sides.   This  statement  is  a  theorem  to  be  proved  in  geometiy. 

49.  Problem.  A  construction  to  be  made  so  that  it  shall 
satisfy  certain  given  conditions  is  called  a  prohlem. 

For  example,  required  to  construct  a  triangle  all  of  whose  sides  shall 
be  equal.  This  construction  was  made  in  §  31,  Ex.  4,  and  later  it  will  be 
proved  that  the  construction  was  correct. 

50.  Proposition.  A  statement  of  a  theorem  to  be  proved  or 
a  problem  to  be  solved  is  called  a  proposition. 

In  geometry,  therefore,  a  proposition  is  either  a  theorem  or  a  problem. 
We  shall  find  that  most  of  the  propositions  at  first  are  theorems.  After 
we  have  proved  a  number  of  theorems  so  that  we  can  prove  that  the 
solutions  of  problems  are  correct,  we  shall  solve  some  problems. 

51.  Corollary.  A  truth  that  follows  from  another  with  little 
or  no  proof  is  called  2,  corollary. 

For  example,  since  we  admit  that  all  straight  angles  are  equal,  it  follows 
as  a  corollary  that  all  right  angles  are  equal,  since  a  right  angle  is  half 
of  a  straight  angle. 


22  PLANE  GEOMETRY 

52.  Axioms.  The  following  are  the  most  important  axioms 
used  in  geometry: 

1.  If  equals  are  added  to  equals  the  sums  are  equal. 

2.  If  equals  are  subtracted  from  equals  the  remainders  are  equal. 

3.  If  equals  are  m^ultvplied  by  equals  the  products  are  equal. 

4.  If  equals  are  divided  by  equals  the  quotients  are  equal. 
In  division  the  divisor  is  never  zero, 

5.  Like  powers  or  like  p)ositive  roots  of  equals  are  equal. 
We  learn  from  algebra  that  the  square  root  of  4  is  +  2  or  —  2,  but  of 

course  these  are  not  equal.    In  geometry  we  shall  use  only  the  positive  roots. 

6.  If  unequals  are  operated  on  by  positive  equals  in  the  same 
way,  the  results  are  unequal  in  the  same  order. 

Taking  a>h  and  taking  x  and  y  as  equal  positive  quantities,  this 
axiom  states  that 

a4-x>&  +  2/,     a  —  x>b  —  y,     ax>by,     ->-,  etc. 

X     y 

7.  If  unequals  are  added  to  unequals  in  the  same  order,  the 
sum,s  are  unequal  in  the  same  order  ;  if  unequals  are  subtracted 
from  equals  the  remainders  are  unequal  in  the  reverse  order. 

If  a  >  6,  c  >  d,  and  x  =  y,  then  a  +  c>h  -^  d,  and  x  —  a  <y  —  h. 

8.  Quantities  that  are  equal  to  the  same  quantity  or  to  equal 
quantities  are  equal  to  each  other. 

9.  A  quantity  may  be  substituted  for  its  equal  in  an  equation 
or  in  an  inequality. 

Thus  if  X  =  &  and  \i  a  -{-  x  =  c,  then  a  +  6  =  c  ;  and  if  a  +  x  >  c,  then 
a  +  6  >  c.  Axiom  8  is  used  so  often  that  it  is  stated  separately,  although 
it  is  really  included  in  Axiom  9. 

10.  If  the  first  of  three  quantities  is  greater  than  the  second, 
and  the  second  is  greater  than  the  third,  then  the  first  is  greater 
than  the  third. 

Thus  if  a  >  6,  and  if  6  >  c,  then  a>  c. 

11.  The  whole  is  greater  than  any  of  its  parts,  and  is  equal 
to  the  sum  of  all  of  its  pjarts. 


INTEODUCTION  23 

53.  Postulates.  The  following  are  among  the  most  impor- 
tant postulates  used  in  geometry.  Others  will  be .  introduced 
as  needed. 

1.  One  straight  line  and  only  one  can  hfi  drawn  through  two 
given  points. 

2.  A  straight  line  may  be  produced  to  any  required  length. 

To  produce  A  B  means  to  extend  it  through  B ;      

to  produce  BA  means  to  extend  it  through  A. 

3.  A  straight  line  is  the  shortest  path  between  two  p)oints. 

4.  A  circle  may  be  described  with  any  given  j^oint  as  a  center 
and  any  given  line  as  a  radices. 

5.  Any  figure  may  be  7noved  from  one  2jl(^(^(i  to  another  with- 
out altering  its  size  or  shajje. 

6.  All  straight  angles  are  equal. 

54.  Corollary  1.    Ttvo  p>oints  determine  a  straight  line. 
This  is  only  a  brief  way  of  stating  Postulate  1. 

55.  Corollary  2.  Two  straight  lines  can  intersect  in  only 
one  point. 

For  if  they  had  two  points  in  common  they  would  coincide  (Post.  1). 

56.  Corollary  3.    All  right  angles  are  equal. 

For  all  straight  angles  are  equal  (Post.  6),  and  a  straight  angle  (§  34) 
is  twice  a  right  angle.    Hence  Axiom  4  applies. 

57.  Corollary  4.  From  a  given  point  in  a  given  line  only 
one  perpendicular  can  be  drawn  to  the  line.  C  b 

For  if  there  could  be  two  perpendiculars 
to  DA  at  0,  as  OB  and  OC,  we  should  have 
angles  AOB  and  AOC  both  right  angles,  which 
is  impossible  (§  56).  j) 


58.  Corollary  5.    Equal  angles  have  equal  complements, 
equal  supplements,  and  equal  conjugates. 

59.  Corollary  6.    The  greater  of  two  angles  has  the  less 
complement,  the  less  suj^plcTnent,  and  the  less  conjugate. 


24  PLANE  GEOMETRY 

EXERCISE  4 

1.  If  10°  +  Z  ic  =  27°  30',  find  the  value  of  Z  x. 

2.  If  Z  a;  +  37°  =  I  Z  ic  +  40°,  find  the  value  of  Z  x. 

3.  If  |Za:  +  Z^>  =  5Z^,  find  the  value  of  Zic. 

4.  lfZ.x-\-/-a  =  ^/.a  —  Z.X,  find  the  value  of  Z ic. 

^mc?  the  value  of  /.x  in  each  of  the  following  equations  : 

5.  Za;+13°  =  39°.  10.  Zic  =  0.7Z;i;  +  33°. 

6.  Zee -17°  =  46°.  11.  Za^  =  0.1Zic+18°.     ^ 

7.  2  Zee -ZicH- 23°.  12.  fZec  =  ^Zec +  2^°. 

8.  5Zx  =  2Zx-4-21°.  13.  |  Zee  =  0.1  Zee +14°. 

9.  4Zee  =  iZee+70°.  14.  f  Za^  =  JZee  +  2°. 

15.  12Za^+17°=9Z.r  +  32°. 

16.  5Za;-22°30'  =  2Zec+ll°. 

17.  51°20'-§Z:r  =  5°l'  +  3Za;. 

18.  73°  21'  4"  -  Z  X  =  3°  3'  12"  +  4  Z  ee. 

19.  If  ec  +  20°  =  y  and  ?/  —  5°  =  2  ee,  what  is  the  value  of  ec 
and  of  ?/  ? 

Find  the  value  of  x  and  of  y  in  each  of  the  following  sets 
of  equations  : 

20.  ee  +  ?/  =  45°,  23.  ^  +  2^  =  21°, 
x-tj  =  S5°.  ^  +  32/  =  26°15'. 

21.  x-Stj  =  0'',  24.      ee  +  2/ =  9°  20' 15", 
a;  +  8?/=:80°.  2ee-2/  =  12°25' 15". 

22.  2ee  +  2/  =  64°,  25.      ee  -     2/  =  5'5", 

See  -  2/  =  88°.  3ee  +  4 1/  =  14°  50'  50". 

26.  If  ec  <  10°  and  y  =  7°  30',  what  can  be  said  as  to  the 
value  oi  x-i-y? 

27.  In  Ex.  26,  what  can  be  said  as  to  the  value  oi  x  —  y? 


BOOK  I 

RECTILINEAR  FIGURES 

Proposition  I.    Theorem 
60.  If  tioo  lines  intersect,  the  vertical  angles  are  equal. 


D^  A 

Given  the  lines  AC  and  BD  intersecting  at  O. 
To  prove  that  AAOB  =  Z.COD. 

Proof.  Z  .1  OB-\-A  BOC  =  a  st.  Z.  §  43 

( The  two  adjacent  angles  which  one  straight  line  makes  with  another 
are  together  equal  to  a  straight  angle.) 

Likewise    Z50C -|-Z  COT)  =  a  st.  Z.  §43 

.-.ZAOB  +  Z  BOC  =  Z  BOC  -\-  Z COD.  Post.  6 

(All  straight  angles  are  equal.) 

.\ZAOB  =  ZCOD.  Ax.  2 

(If  equals  are  subtracted  from  equals  the  remainders  are  equal.)   Q.E.D. 

61.  Nature  of  a  Proof.  From  Prop.  I  it  is  seen  that  a  tlieorem 
has  (1)  certain  things  given;  (2)  a  definite  thing  to  he p)roved ; 
(3)  a  proof,  consisting  of  definite  statements,  each  supported 
by  the  authority  of  a  definition,  an  axiom,  a  postulate,  or  some 
proposition  previously  proved. 

25 


26  BOOK  I.    PLANE  GEOMETEY 

62.  Triangles  classified  as  to  Sides.    A  triangle  is  said  to  be 

scalene  when  no  two  of  its  sides  are  equal ; 
isosceles  when  two  of  its  sides  are  equal ; 
equilateral  when  all  of  its  sides  are  equal. 


Scalene  Isosceles  Equilateral 

63.  Triangles  classified  as  to  Angles.    A  triangle  is  said  to  be 

right  when  one  of  its  angles  is  a  right  angle ; 
obtuse  when  one  of  its  angles  is  an  obtuse  angle ; 
acute  when  all  of  its  angles  are  acute  angles ; 
equiangular  when  all  of  its  angles  are  equal. 


Right  Obtuse  Acute  Equiangular 

64.  Corresponding  Angles  and  Sides.  If  two  triangles  have 
the  angles  of  the  one  respectively  equal  to  the  angles  of  the 
other,  the  equal  angles  are  called  corresponding  angles,  and  the 
sides  opposite  these  angles  are  called  corresponding  sides. 

Corresponding  parts  are  also  called  homologous  parts. 

65.  Square.  A  rectilinear  figure  having  four  equal  sides  and 
four  right  angles  is  called  a  square. 

66.  Congruent.  If  two  figures  can  be  made  to  coincide  in  all 
their  parts,  they  are  said  to  be  congruent. 

67.  Corollary.     Corresponding  parts  of  congruent  figures 

are  equal. 

When  equal  figures  are  necessarily  congruent,  as  in  the  case  of  angles 
or  straight  lines,  the  word  equal  is  used.    For  symbols  see  page  vi. 


TKIANGLES  27 

Proposition  II.    Theorem 

68.  Tivo  triangles  are  congruent  if  tivo  sides  and  the 
included  angle  of  the  one  are  equal  respectively  to  tivo 
sides  and  the  included  angle  of  the  other. 


Given  the  triangles  ABC  and  XYZ,  with  AB  equal  to  XYy  AC 
equal  to  XZ,  and  the  angle  A  equal  to  the  angle  X 

To  prove  that     A  ABC  is  congruent  to  AXYZ. 

Proof.    Place  the  A  ABC  upon  the  AXYZ  so  that  A  shall 
fall  on  X  and  AB  shall  fall  along  AT.  Post.  5 

{Any  figure  may  be  moved  from  one  place  to  another  without 
altering  its  size  or  shape.) 

Then  B  will  fall  on  F, 
{For  AB  is  given  equal  to  XY.) 

^C  will  fall  along  XZ, 
{For  ZA  is  given  equal  to  ZX.) 

and  C  will  fall  on  Z. 
{For  AC  is  given  equal  to  XZ.) 

.'.  CB  will  coincide  with  ZY.  Post.  1 

{One  straight  line  and  only  one  can  he  drawn  through  two  given  points.) 

.'.  the  two  A  coincide  and  are  congruent,  by  §  66.     q.e.d. 

69.  Corollary.    Two  right  triangles  are  congruent  if  the 
sides  of  the  right  angles  are  equal  respectively. 

The  right  angles  are  equal  {§  56).    How  does  Prop.  II  apply  ? 


28 


BOOK  I.    PLANE  GEOMETEY 


EXERCISE  5 

1.  In  this  figure  if  Z  a  =  53°,  how  many  degrees  are  there 
inZ^/?  inZcc?  inZ^? 

2.  In  Ex.  1,  if  Z  a  were  increased  to  89°,  what        ^^^ 
would  then  be  the  size  of  A  x,  y,  and  z  ? 

3.  In  the  square  ABCD,  prove  that  ^C  =  J5Z). 

In  ^ABC  and  BAB  what  two  sides  of  the  one  are 
known  to  be  equal  to  what  two  sides  of  the  other  ? 
How  about  the  included  angles  ?  "Write  a  complete 
proof  as  in  Prop.  II. 

4.  If  ABCD  is  a  square  and  P  is  the  mid- 
point of  AB,  prove  that  PC  =  PD. 

What  triangles  should  be  proved  congruent  ?  Can 
this  be  done  by  Prop.  II  ?    Write  the  proof. 

5.  How  many  degrees  in  an  angle  that  equals    a  .      p         b 
one  fourth  of  its  complement  ?  one  tenth  of  its  complement  ? 

6.  How  many  degrees   in  an  angle   that  equals   twice   its 
supplement  ?    one  third  of  its  supplement  ? 

7.  In  the  square  ABCD  the  points  P,  Q, 
i?,  5  bisect  the  consecutive  sides.  Prove  that 
PQ=::QR  =  RS=SP. 

8.  In  the  square  ABCD  the  point  P  bisects 
CD,  and  BM  is  made  equal  to  AN,  as  shown  in 
this  figure.    Prove  that  PM  =  PN. 

What  two  sides  and  included  angle  of  one  triangle 
must  be  proved  equal  to  what  two  sides  and  included  N 
angle  of  another  triangle  ? 

9.  Prove  that  to  determine  the  distance  AB 
across  a  pond  one  may  sight  from  A  across  a 
post  P,  place  a  stake  at  A'  making  PA'  =  AP, 
then  sight  along  BP  making  PB' =  BP,  and 
finally  measure  A'B'.  ^^ 


TRIANGLES 


29 


70.  Drawing  the  Figures.  Directions  have  already  been  given 
(§  31)  for  drawing  the  most  common  geometric  figures.  For 
example,  in  Prop.  II  the  complete  work  of  drawing  AXYZ 
so  that  XY=AB,  Z.X  =  /.A,  and  XZ  =  AC,  is  indicated  in 
the  following  figures,  the  construction  lines  being  dotted^  as  is 
always  the  case  in  this  book. 


It  is  desirable  to  construct  such  figures  accurately,  employing  com- 
passes and  ruler  until  such  time  as  the  use  of  these  instruments  is 
thoroughly  understood.  Eventually,  however,  the  figures  should  be 
rapidly  but  neatly  draw^n,  free-hand  or  with  the  aid  of  the  ruler,  as 
the  mathematician  usually  makes  his  figures. 

71.  Designating  Corresponding  Sides  and  Angles.  It  is  helpful 
in  propositions  concerning  equality  of  figures  to  check  the  equal 
parts  so  that  the  eye  can  follow  the  proof  more  easily.  Thus  it 
would  be  convenient  to  represent  the  above  figures  as  follows : 

C  Z 


A  •  B  X  Y 

Here  AB  and  XFhave  one  check,  AC  and  XZ  two  checks,  and  the 
equal  angles  A  and  X  are  marked  by  curved  arrows. 

If  a  figure  is  very  complicated,  there  is  sometimes  an  advan- 
tage in  using  colored  crayons  or  colored  pencils,  but  otherwise 
this  expedient  is  of  little  value. 

While  such  figures  have  some  attraction  for  the  eye  they  are  not  gen- 
erally used  in  practice,  one  reason  being  that  the  student  rarely  has  a 
supply  of  colored  pencils  at  hand  when  studying  by  himself. 


80  BOOK  I.    PLANE  GEOMETRY 

Proposition  III.    Theorem 

72.  Two  triangles  are  congruent  if  two  angles  and 
the  included  side  of  the  one  are  equal  respectively  to 
tivo  angles  and  the  included  side  of  the  other. 


T 

Given  the  triangles  ABC  and  XYZ,  with  angle  A  equal  to  angle  X, 
angle  B  equal  to  angle  F,  and  with  AB  equal  to  XY. 

To  prove  that     A  ABC  is  congruent  to  AXYZ. 

Proof.  Place  the  A  ABC  upon  the  AXYZ  so  that  AB  shall 
coincide  with  its  equal,  XY.  Post.  5 

{Any  figure  may  be  moved  from  one  place  to  another  without 
altering  its  size  or  shape.) 

Then  AC  will  fall  along  XZ  and  BC  along  YZ. 
{For  it  is  given  that  ZA  =  ZX  and  ZB  =  ZY.) 

.'.  C  will  fall  on  Z.  §  55 

{Two  straight  lines  can  intersect  in  only  one  point.) 

.'.  the  two  A  are  congruent.  §  66 

{If  two  figures  can  he  made  to  coincide  in  all  their  parts,  they 

are  said  to  be  congruent.)  Q.E.D. 

73.  Hypothesis.  A  supposition  made  in  an  argument  is  called 
an  h7/j)othesls. 

Thus,  where  it  is  said  that  ZA  =  ZX  and  ZB  =  ZY,  yve  might  say- 
that  this  is  true  "by  hypothesis,"  instead  of  saying  that  Z^  is  given 
equal  to  Z  X,  and  Z  jB  is  given  equal  to  Z  Y.  The  word  is  generally 
used,  however,  for  an  assumption  made  somewhere  in  the  proof. 


TRIANGLES 


31 


M 


EXERCISE  6 

1.  In  the  square  ABCD  the  point  7^  bisects  CD,  and  PQ  and 
PR  are  drawn  so  that  Z  QPC  =  30°  and  Z  RPQ  =  120°.  Prove 
that  PQ  =  PR. 

If  ZQPC  =  30°  and  ZRPQ  =  120°,  what  does  ZDPR 
equal  ? 

In  the  two  triangles  what  parts  are  respectively- 
equal,  and  why  ? 

Write  the  proof  in  full.  A 

2.  In  this  figure  prove  that  if  CAT  bisects  Z  .1 CB  and  is  also 
perpendicular  to  AB,  the  triangle  ABC  is  isosceles.  ^ 

In  i^AMC  and  BMC  are  two  angles  of  the  one  respec- 
tively equal  to  two  angles  of  the  other  ?   Why  ? 
The  two  triangles  have  one  common  side. 
Write  the  proof  in  full. 

3.  In  the  triangle  ABC,  AC  =  BC  and  CM  bisects  the  angle  C. 
Prove  that  CM  bisects  the  base  AB. 

4.  The  triangle  ABC  has  Z.A  equal  to  Z.B. 
The  point  P  bisects  AB,  and  the  lines  PM 
and  PN  are  drawn  so  that  Z.BPM=Z.NPA. 
Prove  t\\2it  BM  =  AN. 

5.  The  triangle  ABC  has  Z  .1  =Z.B.  The  lines 
AP  and  BQ  are  so  drawn  that  ZBAP  =  Z.QBA. 
Prove  that  AP  =  BQ. 

6.  Wishing  to  measure  the  distance  across  a  river,  some 
boys  sighted  from  A  to  a  point  P.  p 
They  then  turned  and  measured  AB 
at  right  angles  to  AP.  They  placed 
a  stake  at  0,  halfway  from  A  to  B, 
and  drew  a  perpendicular  to  AB  at  B. 
They  placed  a  stake  at  C,  on  this 
perpendicular,  and  in  line  with  0  and  P.  They  then  found 
the  width  by  measuring  BC.    Prove  that  they  were  right 


32  BOOK  I.    PLANE  GEOMETRY 

Proposition  IV.    Theorem 

74.  In  an  isosceles  triangle  the  angles  opposite  the 

equal  sides  are  equal. 

c 


Given  the  isosceles  triangle  ABC^  with  AC  equal  to  BC. 

To  prove  that  Z.A  =  /.B. 

Proof.    Suppose  CD  drawn  so  as  to  bisect  Z  A  CB. 
Then  in  the  A  ADC  and  BDC, 

AC  =  BC,  Given 

CD  =  CD,  Iden. 

{That  is,  CD  is  common  to  the  two  triangles.) 

and  ZACD  =  ZDCB.  Hyp. 

{For  CD  bisects  Z  AC B.) 

.-.A  ADC  is  congruent  to  A  BDC.  §  68 

( Two  A  are  congruent  if  two  sides  and  the  included  Z  of  the  one  are  equal 
respectively  to  two  sides  and  the  included  Z  of  the  other.) 

,\ZA=ZB.  §67 

{Corresponding  parts  of  congruent  figures  are  equal.)        Q.  E.  D. 

This  proposition  has  long  been  known  as  the  Pons  asinorum,  or  Bridge 
of  Fools  (asses).    It  is  attributed  to  Thales,  a  Greek  philosopher. 

In  an  isosceles  triangle  the  side  which  is  not  one  of  the  two  equal 
sides  is  called  the  base. 

75.  Corollary.   An  equilateral  triangle  is  equiangular. 
Is  ah  equilateral  triangle  a  special  kind  of  isosceles  triangle  ? 


tria:kgles  33 

EXERCISE  7 

1.  With  the  figure  of  Prop.  IV,  if  AC  =  BC  and  CD  bisects 
Z  C,  prove  that  CD  is  ±  to  AB. 

What  angles  must  be  proved  to  be  right  angles  ?  What 
is  a  right  angle  ?  Do  these  angles  fulfill  the  require- 
ments of  the  definition  ? 

2.  In  the  adjacent  figure  ^C  =  -BC.    Prove  that    ^ 
/-in=^  Z.  n. 

3.  In  the  following  figure  AC  =  BC  and  AD  =  BD.    Prove  that 
Z.CBD=^ADAC. 

What  angles  are  equal  by  Prop.  IV  ?  Then  what 
axiom  applies  ? 

4.  In  the  figure  of  Ex.  3  prove  that  if  a  line    ^ 
is  drawn  from  C  to  D,  the  A  DBC  is  congruent 
to  the  A  DA  C.  b 

5.  Two  isosceles  triangles,  ABC  and  ABD,  are  constructed 
on  the  same  side  of  the  common  base  AB.    Prove 

thsA.  Z.CBD  =  Z.DAC. 


c 


6.  In  the  figure  of  Ex.  5  prove  that  a  line 
drawn  through  C  and  D  bisects  Z  ADB. 

What  two  triangles  must  be  proved  congruent  ? 

7.  Inthisfigure.4C  =  5Cand^P=BQ.    Prove  that  PC  =QC. 
Also  prove  that  Z  MPC  =  Z  CQM.  C 

8.  In  this  figure,  if  AC  =  BC,  AP  =  BQ, 
and  PM=QM,  prove  that  CM  is  ±  to  PQ. 

What  angles  must  be  proved  to  be  right  angles  ?       J-^ -^ Q~^B 

9.  In  this  figure  P,  Q,  and  R  are  mid-points  of  the  sides  of 
the  equilateral  triangle  ABC.    Prove  that  PQR  is  ^ 
an  equilateral  triangle.                                                     ^^     x  g 

Prove  that  /kAPR,  BQP,  and  CRQ  are  congruent  by 
using  two  propositions  already  proved.  a 


34  BOOK  I.    PLANE  GEOMETEY 

Proposition  V.    Theorem 

76.  If  hvo  angles  of  a  triangle  are  equal,  the  sides 
opposite  the  equal  angles  are  equal,  and  the  triangle 
is  isosceles. 


c  c' 


A'  B' 

Given  the  triangle  ABC,  with  the  angle  A  equal  to  the  angle  B. 
To  prove  that  AC  =  BC. 

Proof.  Suppose  the  second  triangle  A'B'C'  to  be  an  exact 
reproduction  of  the  given  triangle  ABC. 

Turn  the  triangle  A'B'C'  over  and  place  it  upon  ABC  so 
that  B'  shall  fall  on  A  and  A'  shall  fall  on  B.  Post.  5 

Then  B'A'  will  coincide  with  AB.  Post.  1 

Since                                    ZA'  =  ZB',  Given 

and                                       ZA=ZA',  Hyp. 

.'.ZA=ZB'.  Ax.  8 

.'.^'C  will  lie  along  .1(7. 

Similarly  A'C  will  lie  along  BC. 

Therefore  C  will  fall  on  both  AC  and  BC,  and  hence  at 

their  intersection.  .    . 

.'.B'C'  =  AC. 

But  B'C'  was  made  equal  to  BC, 

.'.  AC  =  BC,  by  Ax.  8.  q.e.d. 

77.  Corollary.    An  equiangular  triangle  is  equilateral. 


TRIANGLES  35 

78.  Kinds  of  Proof.  In  the  five  propositions  thus  far  proved 
in  the  text  two  different  kinds  of  proof  have  been  seen : 

(1)  Synthetic.  In  Prop.  I  we  put  together  some  known  truths 
in  order  to  obtain  a  new  truth.  Such  a  method  of  proof  is  known 
as  the  synthetic  method,  and  is  the  most  common  of  all  that  are 
used  in  geometry. 

In  this  method  we  endeavor  simply  to  find  what  propositions 
have  already  been  proved  that  will  lead  to  the  proof  of  the 
proposition  that  is  before  us.  This  method  was  used  in  all  the 
exercises  on  pages  28,  31,  and  33. 

(2)  By  superposition.  In  Props.  II  and  III  we  placed  one  figure 
on  another  and  then,  by  synthetic  reasoning,  showed  them  to 
be  identically  equal.  Such  proof  is  known  as  a  proof  by  super- 
position. Superposition  means  "  placing  on,"  and  one  figure  is 
said  to  be  superposed  on  the  other. 

In  Prop.  V  a  special  kind  of  proof  by  superposition  was 
employed,  in  which  we  superpose  a  figure  on  its  exact  dupli- 
cate. This  special  method  is  rarely  used,  but  in  this  proposi- 
tion it  materially  simplifies  the  proof. 

79.  Converse  Propositions.  If  two  propositions  are  so  related 
that  what  is  given  in  each  is  what  is  to  be  proved  in  the  other, 
each  proposition  is  called  the  converse  of  the  other. 

E.g.  in  Prop.  IV  we  have  given 

AC  =  BC,  to  prove  that  ZA  =  ZB. 
In  Prop.  V  we  have  given 

ZA  =  ZB,  to  prove  that  AC  =  BC. 

Hence  Prop.  V  is  tlie  converse  of  Prop.  IV,  and  Prop.  IV  is  the  con- 
verse of  Prop.  V. 

Not  all  converses  are  true,  and  hence  we  have  to  prove  any 
given  converse. 

E.g.  tlie  converse  of  the  statement  ''Two  right  angles  are  two  equal 
angles"  is  "Two  equal  angles  are  two  right  angles,"  and  this  statement 
is  evidently  false. 


36  BOOK  I.  pla:^^e  geometry 

Proposition  VI.    Theorem 

80.  Tioo  triangles  are  congruent  if  the  three  sides  of  the 
one  are  equal  respectively  to  the  three  sides  of  the  other. 


Given  the  triangles  ABC  and  A'B'C\  with  AB  equal  to  A^B^ 
AC  equal  to  A'C,  and  BC  equal  to  B'C 

To  prove  that     A  ABC  is  congruent  to  AA'B'C. 

Proof.    Let  AB  and  A'B'  be  the  greatest  of  the  sides  of  the  A. 

Place  A  A'B'C  next  to  A  ABC  so  that  A'  shall  fall  on  A, 
the  side  A'B'  shall  fall  along  AB,  and  the  vertex  C'  shall  be 
opposite  the  vertex  C.  Post.  5 

Then  B'  will  fall  on  B. 

{For  A'B''  is  given  equal  to  AB.) 

Draw  CC. 

Since                                  AC  =  AC',  Given 

.•.ZACC'  =  ZCCA.  §74 

Since                                 BC=BC',  Given 

.'.ZC'CB  =  ZBC'C.  §74 

.-.  Z  .1  CC"+  Z  C'CB  =  Z  CCA  +  Z BC'C.  Ax.  1 

Hence                        ZylC5  =  Z5C"^.  Ax.  11 

{For  Z  A CB  is  made  up  of  ZACC  and  Z C'CB,  and  A  BC'A  is  made 
up  of  Z  CCA  and  Z  BC'C.) 

.-.  A  ABC  is  congruent  to  A  ABC'.  §  68 

.*.  A.4J5C  is  congruent  to  A.l'^'C',  by  Ax.  9.       q.e.d. 


TRIANGLES 


37 


EXERCISE  8 

1.  Prove  that  a  line  from  the  vertex  to  the  mid-point  of 
the  base  of  an  isosceles  triangle  cuts  the  triangle  into  two 
congruent  triangles. 

2.  Three  iron  rods  are  hinged  at  the  ex- 
tremities, as  shown  in  this  figure.  Is  the 
figure  rigid  ?    Why  ? 

3.  Four  iron  rods  are  hinged,  as  shown  in 
this  figure.  Is  the  figure  rigid  ?  If  not,  where 
would  you  put  in  the  fifth  rod  to  make  it  rigid  ? 
Prove  that  this  would  accomplish  the  result. 

4.  If  two  isosceles  triangles  are  constructed  on  opposite 
sides  of  the  same  base,  prove  by  Prop.  VI  and  §  58  that 
the  line  through  the  vertices  bisects  the 
vertical  angles. 

5.  In  this  figure  AB  =  AD  and  CB  =  CD.  ^' 
Prove  that  AC  bisects  Z  BAD  and  Z  DCB. 

6.  In  §  31,  Ex.  8,  it  was  shown  how  to  bisect 
an  angle,  this  being  the  figure  used.  Draw  PX 
and  PY,  and  prove  by  Prop.  VI  that  PO  bi- 
sects ZAOB. 

7.  In  a  triangle  ABC  it  is  known  that  A  C  =  BC. 
It  A  A  and  Z  B  are  both  bisected  by  lines  meet- 
ing at  P,  prove  that  AABP  is  isosceles. 

8.  In  this  figure  it  is  known  that  Am  =  A n. 
Prove  that  ^C  =  ^C. 

9.  From   the   vertices  A  and  B  of  an  equilateral  triangle 
lines  are  drawn  to  the  mid-points  of  the  opposite  c 
sides.    Prove  that  these  two  lines  are  equal. 

In    A  ABQ  and   BAP  show   that   the    conditions  of 
congruence  as  stated  in  Prop.  II  are  fulfilled. 


38  BOOK  L    PLANE  GEOMETRY 

Proposition  VII.    Theorem 

81.  The  sum  of  tivo  lines  from  a  given  point  to  the 
extremities  of  a  given  line  is  greater  than  the  sum  of 
tivo  other  lines  similarly  drawn,  hut  included  hy  them. 


B 

Given  CA  and  C5,  two  lines  drawn  from  the  point  C  to  the 
extremities  of  the  line  AB^  and  PA  and  PB  two  lines  similarly 
drawn,  but  included  by  CA  and  CB. 

To  prove  that        CA-\-CB>PA-}-  PB. 

Proof.    Produce  AP  to  meet  the  line  CB  at  Q.  Post.  2 

Then  CA -\- CQ>PA  ^ PQ.  Post.  3 

{A  straight  line  is  the  shortest  path  between  two  points.) 

Likewise  .BQ-\-PQ>PB.  Post.  3 

Add  these  inequalities,  and  we  have 

CAJrCQ  +  BQ  +  PQ>PA+PQ^PB.  Ax.  7 

{If  unequals  are  added  to  unequals  in  the  same  order,  the  sums  are  unequal 
in  the  same  order.) 

Substituting  for  CQ  +  BQ  its  equal  CB,  we  have 

CA-\-CB-i-  PQ  >PA  +PQ  +  PB.  Ax.  9 

{A  quantity  may  he  substituted  for  its  equal  in  an  equation  or  in  an 
inequality.) 

Taking  PQ  from  each  side  of  the  inequality,  we  have 

CA^CB>PA-\- PB,  by  Ax.  6.  q. e. d. 


TEIANGLES  39 

Proposition  VIII.    Theorem 

82.  Only  one  perpendicular  can  he  drawn  to  a  given 
line  from  a  given  external  point. 


Given  a  line  XF,  P  an  external  point,  PO  a  perpendicular  to  XY 
from  P,  and  PZ  any  other  line  from  P  to  XY. 

To  prove  that  PZ  is  not  ±  to  XY. 

Proof.    Produce  PO  to  P',  making  OP'  equal  to  PO.  Post.  2 

Draw  P'Z.  Post.  1 

By  construction  POP'  is  a  straight  line. 

.'.  PZP'  is  not  a  straight  line.  Post.  1 

Hence  Z  P'ZP  is  not  a  straight  angle.  §  33 

Since                  A  POZ  and  ZOP'  are  rt.  A,  §  27 

.'.ZP0Z  =  ZZ0P'.  §56 

Furthermore                       PO  =  OP',  Hyp. 

and                                           OZ  —  OZ.  Iden. 

.-.  A OPZ  is  congruent  to  AOP'Z,  §  68 

( Two  A  are  congruent  if  two  sides  and  the  included  Z  of  the  one  are 
equal  respectively  to  two  sides  and  the  included  Z  of  the  other.) 

and  Z  OZP  =  Z  P'ZO.  §  67 

.*.  Z  OZP,  the  half  of  Z  P'ZP,  is  not  a  right  angle.    §  34 

.-.  PZ  is  not  ±  to  Xr,  by  §  27.  Q.e.d. 


40  BOOK  I.    PLANE  GEOMETEY 

Proposition  IX.    Theorem 

83.  Tioo  lines  draion  from  a  point  in  a  perpendicu- 
lar to  a  given  line,  cutting  off  on  the  given  line  equal 
segments  from  the  foot  of  the  perpendicular,  are  equal 
and  make  equal  angles  ivith  the  perpendicular. 


A  o  B        ^ 

Given  PO  perpendicular  to  XF,  and  PA  and  PB  two  lines  cutting 
off  on  XY  equal  segments  OA  and  OB  from  0. 

To  prove  that  PA  =  PB, 

and  ZAPO  =  ZOPB. 

Proof.    In  the  A  A  OP  and  BOP, 

Z  POA  and  Z  BOP  are  rt.  A.  §  27 

{For  PO  is  given  ±  to  XY.) 

.'.ZP0A=ZB0P.  §56 

{All  right  A  are  equal.) 

Also  OA  =  OB,  Given 

and  PO  =  PO.  Men. 

{That  is,  PO  is  common  to  the  two  ^.) 

.-.A  A  OP  is  congruent  to  A  BOP.  §  68 

( Two  A  are  congruent  if  two  sides  and  the  included  Z  of  the  one  are  equal 
respectively  to  two  sides  and  the  included  Z  of  the  other.) 

.-.  PA  =  PB, 
and  AAPO  =  Z  OPB.  §  67 

{Corresponding  parts  of  congruent  figures  are  equal.)         Q.E.D. 


TRIANGLES 
Proposition  X.    Theorem 


41 


84.  Of  two  lines  draion  from  a  point  in  a  perpen- 
dicular to  a  given  line,  cutting  off  on  the  given  line 
unequal  segments  from  the  foot  of  the  perpendicidar, 
the  more  remote  is  the  greater. 


Given  PO  perpendicular  to  XY,  PA  and  PC  two  lines  drawn 
from  P  to  XY,  and  OA  greater  than  OC. 

To  prove  that  PA>PC. 

Proof.    Take  OB  equal  to  OC,  and  draw  PB. 

Then                                  PB  =  PC.  §  83 
Produce  PO  to  P',  making  OP'  =  PO,  and  draw  P'A  and  P'B. 

Then                    PA  =  P'A  and  PB  =  P'B.  §  83 

But                          PA  +  P'A  >PB  +  P'B.  §  81 

.  • .  2  PA  >2PB  and  PA  >  PB.  Axs.  9  and  6 

.'.PA  >PC,  by  Ax.  9.  q.e.d. 

85.  Corollary.  Onl^  two  equal  obliques  can  be  drawn  from 
a  given  point  to  a  given  line,  and  these  cut  off  equal  segments 
from  the  foot  of  the  perpendicular. 

Of  two  unequal  lines  from  a  point  to  a  line,  the  greater  cuts 

off  the  greater  segment  from  the  foot  of  the  perpendicular. 

For  PB  =  PC,  but  PB  cannot  equal  PA  (§  84).  The  segments  OB 
and  OC  are  equal,  for  otherwise  PB  could  not  equal  PC 


42 


BOOK  I.    PLAKE  GEOMETRY 
Proposition  XI.    Theorem 


86.  The  perpendicular  is  the  shortest  line  that  can  he 
drawn  to  a  given  line  from  a  given  external  point. 


Given  a  line  XF,  P  an  external  point,  PO  the  peri)endicular,  and 
PZ  any  other  line  drawn  from  P  to  XY. 

To  prove  that  PO  <  PZ. 

Proof.    Produce  PO  to  P',  making  OP'  =  PO)  and  draw  P'Z. 

Then  PZ  =  P'Z.  §  83 

{Two  lines  drawn  from  a  point  in  a  1.  to  a  given  line,  cutting  off  on  the 
given  line  equal  segments  from  the  foot  of  the  ±,  are  equal.) 

.•.PZ  +  P'Z  =  2  PZ.  Ax.  1 

Furthermore          PO  +  P'O  =  2  PO.  Ax.  1 

But                         PO  +  P'O  <PZ-\-  P'Z.  Post.  3 

.'.2PO<2PZ.  Ax.  9 

.'.  PO<PZ,  by  Ax.  6.  q.e.d. 

87.  Hypotenuse.    The  side  opposite  the  right  angle  in  a  right 
triangle  is  called  the  hypotenuse. 

The  other  two  sides  of  a  right  triangle  are  usually  called  the  sides. 

88.  Distance.    The  length  of  the  straight  line  from  one  point 
to  another  is  called  the  distance  between  the  points. 

The  length  of  the  perpendicular  from  an  external  point  to  a 
line  is  called  the  distance  from  the  point  to  the  line. 


TRIANGLES  43 

Proposition  XII.    Theorem 

89.  Tico  right  triangles  are  congruent  if  the  hypote- 
nuse and  a  side  of  the  one  are  equal  respectively  to  the 
hypotenuse  and  a  side  of  the  other. 

A 


/ 


Given  the  right  triangles  ABC  and  A'5'C',  with  the  h3rpotenuse 
AC  equal  to  the  hypotenuse  A^C\  and  with  BC  equal  to  5'C'. 

To  prove  that     A  ABC  is  congruent  to  AA'B'C. 

Proof.  Place  A  ABC  next  to  A.4'^'C',  so  that  BC  shall  fall 
along  B'C,  B  shall  fall  on  B'j  and  A  and  A '  shall  fall  on  opposite 


sides  of  5'C'. 
Then 

and 

Since 


C  will  fall  on  C, 
{For  BC  is  given  equal  to  B'C\) 

BA  will  fall  along  A  'B'  produced. , 
{For  A  CBA  +  Z  A'B'C'=  a  st.  Z.) 

AC'  =  A'C\ 


Post.  5 


§34 


.•.AB'  =  A'B'.  §85 

.-.  A  ABC  is  congruent  to  A  A^B^C.  §  80 

( Two  A  are  congruent  if  the  three  sides  of  the  one  are  equal  respectively 
to  the  three  sides  of  the  other.)  Q.  E.  D. 

90.  Corollary.  Two  right  triangles  are  congruent  if  any 
two  sides  of  the  one  are  equal  respectively  to  the  corresponding 
two  sides  of  the  other. 


44 


BOOK  I.    PLANE  GEOMETRY 


EXERCISE  9 

1.  ABCD  is  a  square  and  Mis  the  mid-point  of  AB.  With 
ikf  as  a  center  an  arc  is  drawn,  cutting  5C  at  P  and  AD  Sit  Q. 
Prove  that  A  MBP  is  congruent  to  A  MA  Q,  and  d  c 
write  the  general  statement  of  this  theorem 
without  using  letters  as  is  done  here. 

This  would  read,  "If  an  arc  is  drawn,  with  the  mid- 
point of  one  side  of  a  square  as  a  center,  cutting  the      ^       ^ 
sides  perpendicular  to  that  side,  then  the  triangles  cut  off  by,"  etc. 

2.  Draw  a  figure  similar  to  that  of  Ex.  1,  but  take  a  radius 
such  that  the  arc  cuts  BC  produced  at  a  point  above  C,  and 
AD  above  D.    Then  prove  that  A  MBP  is  congruent  to  A  MA  Q. 

3.  Prove  that  if  from  the  point  P  the  perpendiculars  PM, 
PN  to  the  sides  of  an  angle  A  OB  are  equal,  the 
point  P  lies  on  the  bisector  of  the  angle  A  OB.  JVx 
Write    the    general    statement    of    this    theorem 
without  using  letters  as  is  done  here.                      ^  ^ 

4.  Prove  that  if  the  perpendiculars  from  the  mid-point  M  of 
the  base  AB  oi  a,  triangle  ^i^C  to  the  sides  of  the 
triangle  are  equal,  then  AA=A B.  What  then 
follows  as  to  the  sides  A  C  and  BC  ?  Write  the  gen- 
eral statement  of  this  theorem  without  referring 
to  a  special  figure. 

5.  Prove  that  if  the  perpendiculars  from  the  extremities  of 
the  base  of  a  triangle  to  the  other  two  sides  are  equal,  the 
triangle  is  isosceles. 

6.  Suppose  0Y1.0X.  With  0  as  a  center  an  arc  is  drawn 
cutting  OX  at  A  and  OF  at  5.  Then  with  A 
as  a  center  an  arc  is  drawn  cutting  OF  at  P, 
and  with  5  as  a  center  and  the  same  radius 
an  arc  is  drawn  cutting  OX  at  Q.  Prove 
that  OP  =  OQ. 

What  triangles  are  congruent  by  Prop.  XII  ? 


TRIANGLES  45 

Proposition  XIII.    Theorem 

91.  Tico  right  triaiigles  are  congruent  if  the  hypotenuse 
and  an  adjacent  angle  of  the  one  are  equal  respectively 
to  the  hyjjotenuse  and  an  adjacent  angle  of  the  other, 
c 


Given  the  right  triangles  ABC,  A'B'C\  with  the  hypotenuse  AC 
equal  to  the  hypotenuse  A'C,  and  with  angle  A  equal  to  angle  A'. 

To  prove  that     A  ABC  is  congruent  to  AA'B'C. 

Proof.    Place  A  ABC  upon  A  A'B'C  so  that  A  shall  fall  upon 
A'  SiudAC  shall  fall  along  A 'C.  Post.  5 

Then  C  will  fall  onC, 

{For  AC  is  given  equal  to  A'C\) 

and  AB  will  lie  along  A'B'. 

{For  ZA  is  given  equal  to  Z  A'.) 

Then  because  C  falls  on  C", 

and  AB  and  B^  are  rt.  A,  Given 

{Since  the  A  are  given  as  ri.  A.) 

.-.  CB  will  coincide  with  C'B'.  §  82 

{Only  one  perpendicular  can  be  drawn  to  a  given  line  from  a 
given  external  point.) 

.'.  A  ABC  is  congruent  to  A  A'B'C^.  §  66 

{If  two  figures  can  he  made  to  coincide  in  all  their  parts, 

they  are  said  to  he  congruent.)  Q.E.D. 


46  BOOK  I.    PLANE  GEOMETRY 

Proposition  XIV.    Theorem 

92.  TiDO  lines  in  the  same  plane  perpendicular  to  the 
same  line  cannot  meet  hoiveve?'  far  they  are  produced. 

X 


Given  the  lines  AB  and  CD  perpendicular  to  XY  at  A  and  C 
respectively. 

To  prove  that  AB  and  CD  cannot  meet  however  far  they 
are  produced. 

Proof,    li  AB  and  CD  can  meet  if  sufficiently  produced,  we 
shall  have  two  perpendicular  lines  from  their  point  of  meeting 

to  the  same  line. 

But  this  is  impossible.  §  82 

.'.  AB  and  CD  cannot  meet.  q.e.d. 

93.  Parallel  Lines.    Lines  that  lie  in  the  same  plane  and 
cannot  meet  however  far  produced  are  called  parallel  lines. 

94.  Postulate  of  Parallels.    Through  a  given  point  only  one 
line  can  he  drawn  parallel  to  a  given  line. 

As  always  in  such  cases  the  word  line  means  straight  line. 

95.  Corollary  1.    Two   lines  iii  the   same  plane  perpen- 
dicular to  the  same  line  are  parallel. 

96.  Corollary  2.    Two  lines  in  the  same  plane  parallel  to 

a  third  line  are  parallel  to  each  other. 

For  if  they  could  meet,  we  should  have  two  lines  through  a  point 
parallel  to  a  line.    Why  is  this  impossible  ? 


PARALLEL  LINES 


47 


Proposition  XV.    Theorem 

97.  If  a  line  is  j^erpendicular  to  one  of  tivo  parallel 
lines,  it  is  perpendicular  to  the  other  also. 


o 

P 

^"^ 

Given  AB  and  Ci>,  two  parallel  lines,  with  XY  perpendicular  to 
AB  and  cutting  CD  at  P. 

To  prove  that  XY  is  ±  to  CD. 

Proof.    Suppose  MN  drawn  through  P  _L  to  XY. 

Then                              MN  is  II  to  AB.  §  95 

But                                  CD  is  11  to  AB.  Given 

.*.  CD  and  MN  must  coincide.  §  94 

But                                 XF  is  _L  to  MN.  Hyp. 

.'.  AT  is  X  to  CD.  Q.B.D. 

98.  Transversal.    A  line  that  cuts  two  or  more  lines  is  called 
a  transversal  of  those  lines. 

99 .  Angles  made  by  a  Transversal. 
If  XY  cuts  AB  and  CD,  the  angles 
a,  d,  g,  f  are  called  interior  angles ; 
h,  c,  h,  e  are  called  exterior  angles. 

The  angles  d  and  /,  and  a  and  g, 
are  called  alternate-interior  angles ;  c 
the  angles  b  and  h,  and  c  and  e,  are 
called  alternate-exterior  angles. 

The  angles  b  and/,  c  and  g,  e  and  a,  h  and  d,  are  called  exterior- 
interior  angles. 


48  BOOK  I.    PLAN^E  GEOMETRY 

Proposition  XVI.    Theorem 

100.  If  tivo  parallel  lines  are  cut  hy  a  transversal,  the 
alternate-interior  angles  are  equal. 

X 


Given  AB  and  CD,  two  parallel  lines  cut  by  the  transversal  XY 
in  the  points  P  and  Q  respectively. 

To  prove  that  ZAFQ  =  ZDQP. 

Proof.    Through  0,  the  mid-point  of  PQ,  suppose  MN  drawn 
_L  to  CD. 

Then  MN  is  likewise  ±to  AB.  §  97 

{A  line  ±  to  one  of  two  \\s  is  1.  to  the  other.) 

Now  A  PMO  and  QNO  are  rt.  A.  §  63 

{Since  A  OMP  and  ONQ  are  rt.  A.) 

But  Z  POM  =  Z  QON,  §  60 

(If  two  lines  intersect,  the  vertical  A  are  equal.) 

and  OP=OQ.  Hyp. 

{For  0  was  taken  as  the  mid-point  of  PQ.) 

.'.A  PMO  is  congruent  to  A  QNO.  §  91 

( Two  right  A  are  congruent  if  the  hypotenuse  and  an  adjacent  Z  of  the  one 
are  equal  respectively  to  the  hypotenuse  and  an  adjacent  Z  of  the  other.) 

.'.ZAPQ  =  ZDQP.  §67 

{Corresponding  parts  of  congruent  figures  are  equal.)         Q.  E.  D. 


PAKALLEL  LINES  49 

Proposition  XVII.    Theorem 

101.  When  two  lines  in  the  same  plane  are  cut  hy  a 
transversal,  if  the  alternate-interior  angles  are  equal, 
the  two  lines  are  parallel. 


Y 
Given  the  lines  AB  and  CD  cut  by  the  transversal  XY  in  the 
points  P  and  Q  respectively,  so  as  to  make  the  angles  APQ  and 
DQP  equal. 

To  prove  that  AB  is  II  to  CD, 

Proof.    Since  we  do  not  know  that  AB  is  II  to  CD,  let  us 
suppose  MN  drawn  through  P  II  to  CD. 

We  shall  then  prove  that  AB  coincides  with  MN. 
Now  Z  MPQ  =  Z  DQP.  §  100 

{If  two  II  lines  are  cut  by  a  transversal,  the  alt.-int.  A  are  equal.) 

But  Z  APQ  =:^^  DQP.  Given 

.'.ZAPQ=:Z.MPQ.  Ax.  8 

{Quantities  that  are  equal  to  the  same  quantity  are  equal  to  each  other.) 

.'.  AB  and  MN  must  coincide.  §  23 

(De/.  of  equal  angles.) 

But  MN  is  II  to  CD.  Hyp. 

{For  MN  was  drawn  II  to  CD.) 

.'.  AB,  which  coincides  with  MN,  is  II  to  CD.       q.e.d. 
This  proposition  is  the  converse  of  Prop.  XVI,  as  defined  in  §  79. 


50  BOOK  I.    PLANE  GEOMETEY 

Proposition  XVIII.    Theorem 

102.  If  tivo  parallel  lines  are  cut  hy  a  transversal,  the 
exterior-interior  angles  are  equal. 


Y 

Given  AB  and  CD,  two  parallel  lines,  cut  by  the  transversal  XY 
in  the  points  P  and  Q  respectively. 

To  prove  that  Z  BPX  =ADQX. 

Proof.                           Z  BPX  =  Z.APQ.  §  60 

ZAPQ  =  ZDQX.  §100 

.'.  ZBPX  =  ZDQX,  by  Ax.  8.  q.e.d. 

103.  Corollary  1.  When  two  lines  are  cut  hy  a  transversal, 
if  the  exterior-interior  angles  are  equal,  the  lines  are  parallel. 

The  proofs  of  §§  103  and  105  are  similar  to  that  of  §  101. 

104.  Corollary  2.  If  two  parallel  lines  are  cut  hy  a  trans- 
versal, the  two  interior  angles  on  the  same  side  of  the  trans- 
versal are  supplementary. 

105.  Corollary  3.  When  two  lines  are  cut  hy  a  transversal, 
if  two  interior  angles  on  the  same  side  of  the  transversal  are 
supplementary,  the  lines  are  parallel. 

106.  Corollary  4.  If  two  parallel  liries  are  cut  hy  a  trans- 
versal, the  alternate-exterior  angles  are  equal. 


TEIANGLES  51 

Proposition  XIX.    Theorem 

107.  The  smn  of  the  three  angles  of  a  triangle  is  equal 
to  two  right  angles. 


c 


A 

Given  the  triangle  ABC. 

To  prove  that     AA-\-A  B-\-ZC=2  H.  A. 

Proof.    Suppose  BY  drawn  II  to  AC,  and  produce  AB  to  X 

Then  Z  XB  Y-\-Z.YBC  +  Z.  CBA  =  2  rt.  A.  §  34 

{For  a  St.  Z  equals  2  rt.  A.) 

But  AA=AXBY,  §102 

and  AC  =  ZYBC.  §100 

.'.ZA+AB-\-AC  =  2Yt.A,  by  Ax.  9.       -q.e.d. 

108.  Corollary  1.  Tf  two  triangles  have  two  angles  of  the 
one  equal  to  two  angles  of  the  other,  the  third  angles  are  equal. 

109.  Corollary  2.  In  a  triangle  there  can  he  hut  one  right 
angle  or  one  obtuse  angle. 

110.  Exterior  Angle.  The  angle  included  by  one  side  of  a 
figure  and  an  adjacent  side  produced  is  called  an  exterior  angle. 

In  the  above  figure  A  XBC  is  an  exterior  angle,  and  A  A  and  C  are 
called  the  opposite  interior  angles. 

111.  Corollary  3.  An  exterior  angle  of  a  triangle  is  equal 
to  the  sum  of  the  two  opposite  interior  angles,  and  is  therefore 
greater  than  either  of  them. 


62 


BOOK  I.    PLANE  GEOMETRY 


EXERCISE  10 

1.  Show  that  if  we  place  a  draftsman's 
triangle  against  a  ruler  and  draw  A  C,  and 
move  the  triangle  along  as  shown  in  the 
figure  and  draw  A'C,  then  ^C  is  II  to  A'C.    [ 

2.  In  the  next  figure  x  =  60°.    How  many  de- 
grees in  each  of  the  other  seven  angles  ? 

3.  In  the  next  figure  representing  two  pairs 
of  parallel  lines  certain  angles  are  equal.    State 
these  equalities  in  this  form :  a  =  c  =  g  =  e  = 
0  =  •■',  and  give  the  reason  in  each  case. 

4.  In  the  figure  of  Ex.  3  state  ten  pairs 
of  nonadjacent  angles  that  are  supplementary. 
Thus  :  a-\-h=  180°  and  cZ  +  e  ==  180°. 

5.  In  the  triangle  ABC,  AC  =BC  and  DE  is 
drawn  parallel  to  AB.  Prove  that  CD  =  CE. 
Write  a  general  statement  of  the  theorem. 

6.  In  the  next  figure  A  B  is  parallel  to  CD,  and 
/-APQ  is  half  of  Z.QPB.  How  many  degrees  in 
the  various  angles  ? 

7.  If  Z  YQD  =  135°,  how  many  degrees  in  the 
various  angles  ? 

8.  Let  ZDQP  =  x  and  Z.YQD  =  y.  Then  if 
y  —  x  =  100°,  find  the  value  of  x  and  y. 

9.  Let  ZCQY  =  x  and  ZXPA=y. 
the  value  of  x  and  y. 

10.  In  the  next  figure  x  =  72° and  x  =  ^y.    It  is  required  to 
know  if  the  lines  are  parallel,  and  why. 

11.  In  the  figure  of  Ex.  10  suppose  x  =  73°  and 
y  —  x  =  32°.  It  is  required  to  know  if  the  lines 
are  parallel,  and  why. 


Then  if  x  =  ^y,  find 


TRIANGLES  53 

The  three  angles  of  a  triangle  are  x,  g,  and  z.  Find  the 
value  of  0,  given  the  values  of  x  and  y  as  follows  : 

12.  X  =  10°,  y  =  30°.  11.  x  =  37°,  y  =  48°. 

13.  a;  =  20°,  2/ =  20°.  18.  ic  =  63°,  y  =  29°. 

14.  X  =  75°,  y  =  50°.  19.  a;  =75°  29',  y  =  68°  41'. 

15.  X  =  38°,  y  =  76°.  20.  x  =  82°  33',  y  =  75°  48'. 

16.  X  =  49°,  y  =  92°.  21.  ic  =  69°  58',  y  =  82°  49'. 

22.  In  a  certain  right  triangle  one  angle  is  37°.  What  is  the 
size  of  the  other  acute  angle  ? 

23.  In  a  certain  right  triangle  one  angle  is  36°  41'.  What  is 
the  size  of  the  other  acute  angle  ? 

24.  In  a  certain  right  triangle  one  angle  is  29°  48'  56". 
What  is  the  size  of  the  other  acute  angle  ? 

25.  In  a  certain  right  triangle  one  acute  angle  is  two  thirds 
of  the  other.    How  many  degrees  are  there  in  each  ? 

26.  In  a  certain  right  triangle  one  acute  angle  is  twice  as 
large  as  the  other.    How  many  degrees  are  there  in  each  ? 

27.  In  a  certain  right  triangle  the  acute  angles  are  2  x  and 
5  X.    Find  the  value  of  x  and  the  size  of  each  angle. 

28.  In  a  certain  triangle  one  angle  is  twice  as  large  as 
another  and  three  times  as  large  as  the  third.  How  many 
degrees  are  there  in  each  ? 

29.  In  a  certain  isosceles  triangle  one  angle  is  twice  another 
angle.    How  many  degrees  in  each  of  the  three  angles  ? 

30.  In  this  figure  what  single  angle  equals 
a  -\-  G?    To  the  sum  of  what  angles  is  q  equal  ? 
also  r  ?   From  these  relations  find  the  number        y^  ^ 
of  degrees  in  ^9  +  5'  +  r. 

31.  Prove  Prop.  XIX  by  first  drawing  a  parallel  to  AB 
through  C,  instead  of  drawing  BY. 


54  BOOK  L    PLANE  GEOMETRY 

Proposition  XX.    Theorem 

112.  The  sum  of  any  two  sides  of  a  triangle  is  greater 
than  the  third  side,  and  the  difference  hetiveen  any  tivo 
sides  is  less  than  the  third  side. 


A  B 

Given  the  triangle  ABC^  with  AB  the  greatest  side. 

To  prove  that     BC  +  CA  >AB,  and  AB  -BC<  CA. 

Proof.  BC  +  CA  >AB.  Post.  3 

{A  straight  line  is  the  shortest  path  between  two  points.) 
Since  BC-]-CA>AB, 

.■.CA>AB-BC;  Ax.  6 

or,  AB  —  BC<CA.  q.e.d. 

EXERCISE  11 

State  in  what  cases  it  is  possible  to  form  triangles  with  rods 
of  the  following  lengths,  and  give  the  reason  : 

1.  2  in.,  3  in.,  4  in.  4.  7  in.,  10  in.,  20  in. 

2.  3  in.,  4  in.,  7  in.  5.   8  in.,  9^  in.,  18  in. 

3.  6  in.,  7  in.,  9  in.  6.  9f  in.,  lOi  in.,  12i  in. 

7.  In  this  figure  prove  that  AB  -\-BC  >AD  -]-DC. 
Why  is  DB+BODC? 
What  is  the  result  of  adding  AD  to  these  unequals  ? 

8.  How  many  degrees  are  there  in  each 
angle  of  an  equiangular  triangle  ?    Prove  it.       ^  ^   ^ 


TRIANGLES  55 

Proposition  XXI.    Theorem 

113.  If  two  sides  of  a  triangle  are  unequal ,  the  angles 
opjjosite  these  sides  are  unequal,  and  the  angle  opposite 
the  greater  side  is  the  greater. 


A  B 

Given  the  triangle  ABC^  with  BC  greater  than  CA. 
To  prove  that  ZBAOZB. 

Proof.    On  CB  suppose  CX  taken  equal  to  CA. 

Draw  AX.  Post.  1 

Then  A  AXC  is  isosceles.  §  62 

Then  Z  CXA  =  Z  XA  C.  §  74 

(In  an  isosceles  A  the  A  opposite  the  equal  sides  are  equal.) 

But  Z  CXA  >ZB.  §  111 

{An  exterior  Z  of  a  A  is  greater  than  either  opposite  interior  Z.) 

Also  ZBAOZ  XA  C.  Ax.  11 

{For  ZXAC  isa part  of  Z  BA C.) 

Substituting  in  this  inequality  for  Z  XAC  its  equal,  Z  CXA, 
we  have  the  inequality 

Z BA C>Z  CXA .  Ax.  9 


Since 

Z^^OZCA'.l, 

and 

Z.CXA>ZB, 

.'.ZBAOZB. 

Ax.  10 

{If  the  first  of  three  quantities  is  greater  than  the  second,  and  the  second  is 
greater  than  the  third,  then  the  first  is  greater  than  the  third.)  Q.  E.  D. 


56  BOOK  T.    PLANE  GEOMETRY 

Proposition  XXII.    Theorem 

114.  If  two  angles  of  a  triangle  are  unequal,  the  sides 
opposite  these  angles  are  unequal,  and  the  side  opposite 
the  greater  angle  is  the  greater. 

c 


Given  the  triangle  ABC^  with  the  angle  A  greater  than  the  angle  B. 

To  prove  that  BOCA. 

Proof.  Now  BC  is  either  equal  to  CA,  or  less  than  CA,  or 
greater  than  CA. 

But  ii  BC  were  equal  to  CA, 
then  the  Z  A  would  be  equal  to  the  Z.B.  §  74 

{For  they  would  he  A  opposite  equal  sides.) 

And  if  CA  were  greater  than  BC, 
then  the  Z  B  would  be  greater  than  the  Z.A.       §  113 

But  if  CA  is  not  greater  than  BC,  this  is  only  another  way 
of  saying  that  BC  is  not  less  than  CA. 

We  have,  therefore,  two  conclusions  to  be  considered, 

ZA=ZB, 
and  ZA<ZB. 

Both  these  conclusions  are  contrary  to  the  given  fact  that 
the  Z  ^  is  greater  than  the  Z  B. 

Since  BC  cannot  be  equal  to  CA  or  less  than  CA  without 
violating  the  given  condition,  .'.  BOCA.  q.e.d. 

This  proposition  is  the  converse  of  Prop.  XXL 


TRIANGLES 
Proposition  XXIII.    Theorem 


57 


115.  If  tivo  triangles  have  two  sides  of  the  one  equal 
respectively  to  two  sides  of  the  other,  hut  the  included 
angle  of  the  first  triangle  greater  than  the  included 
angle  of  the  second,  then  the  third  side  of  the  first  is 
greater  than  the  third  side  of  the  second. 


Y  Y 

Given  the  triangles  ABC  and  XYZ,  with  CA  equal  to  ZX  and 
BC  equal  to  FZ,  but  with  the  angle  C  greater  than  the  angle  Z. 

To  2)rove  that  AB  >XY. 

Proof.  Place  the  A  so  that  Z  coincides  with  C  and  ZX  falls 
along  CA.  Then  .A'  falls  on  A,  since  ZX  is  given  equal  to  CA, 
and  Z  Y  falls  within  Z  A  CB,  since  AC  is  given  greater  than  Z  Z. 

Suppose  CP  drawn  to  bisect  the  Z  YCB,  and  draw  YP. 


Then  since 

CP  =  CP, 
CY  =  CB, 

Iden. 
Given 

d 

Zycp  =  Zpcb 

} 

Hyp. 

. 

'.  A  PYC  is  congruent 

to  A  PBC. 

§68 

.•.PY=PB. 

§67 

Now 

AP-{-PY>AY. 

.•.AP-^PB>AY. 

.\AB>AY. 

Post.  3 

Ax.  9 

Ax.  11 

.\AB>XY, 

by 

Ax. 

9. 

Q.E.D. 

68  BOOK  I.    PLANE  GEOMETRY 

Proposition  XXIV.    Theorem 

116.  If  tivo  triangles  have  two  sides  of  the  one  equal 

respectively  to  tivo  sides  of  the  other ^  hut  the  third  side 

of  the  first  triangle  greater  than  the  third  side  of  the 

second,  then  the  angle  opposite  the  third  side  of  the  first 

is  greater  than  the  angle  opposite  the  third  side  of 

the  second. 

z 


A  B  X  Y 

Given  the  triangles  ABC  and  XYZ^  with  CA  equal  to  ZX  and  BC 
equal  to  FZ,  but  with  AB  greater  than  XY. 

To  prove  that  the  Z.  C  is  greater  than  the  AZ. 

Proof.    Kow  the  Z  C  is  either  equal  to  the  Z  Z,  or  less  than 
the  Z  Z,  or  greater  than  the  Z  Z. 

But  if  the  Z  C  were  equal  to  the  Z  Z, 

then  the  A  ^5C  would  be  congruent  to  the  A  XFZ,    §  68 

{^or  it  would  have  two  sides  and  the  included  Z  of  the  one  equal  respectively 
to  two  sides  and  the  included  Z  of  the  other.) 

and  AB  would  be  equal  to  XY.  §  67 

And  if  the  Z  C  were  less  than  the  Z  Z, 

then  AB  would  be  less  than  XY.  §  115 

Both  these  conclusions  are  contrary  to  the  given  fact  that 
AB  is  greater  than  XY. 

.'.  AOAZ.  Q.E.D. 

This  proposition  is  the  converse  of  Prop.  XXIII. 


QUADRILATERALS  59 

117.  Quadrilateral.    A  portion  of  a  plane  bounded  by  four 
straight  lines  is  called  a  quadrllateraL 

118.  Kinds  of  Quadrilaterals.    A  quadrilateral  may  be 
a  trapezoid,  having  two  sides  parallel ; 

2^  parallelogram,  having  the  opposite  sides  parallel. 

If  the^nonparallel  sides  are  equal,  a  trapezoid  is  called  isosceles. 
A  quadrilateral  with  no  two  sides  parallel  is  called  a  trapezium. 


Trapezoid  Parallelogram  Tiai)eziuin 

119.  Kinds  of  Parallelograms.    A  parallelogram  may  be 
a  rectangle,  having  its  angles  all  right  angles ; 
a  rhombus,  having  its  sides  all  equal. 

A  parallelogram  with  all  its  angles  oblique  is  called  a  rhomboid. 


Rectangle  Rhombus  Rhomboid 

120.  Base.  The  side  upon  which  a  hgure  is  supposed  to 
rest  is  called  the  base. 

If  a  quadrilateral  has  a  side  parallel  to  the  base,  this  is  called  the 
wpper  base,  the  other  being  called  the  lower  base. 

In  an  isosceles  triangle  the  vertex  formed  by  the  equal  sides  is  taken 
as  the  vertex  of  the  triangle,  and  the  side  opposite  this  vertex  is  taken  as 
the  base  of  the  triangle. 

121.  Altitude.  The  perpendicular  distance  between  the  bases 
of  a  parallelogram  or  trapezoid  is  called  the  altitude. 

The  perpendicular  distance  from  the  vertex  of  a  triangle  to 
the  base  is  called  the  altitude  of  the  triangle. 

122.  Diagonal.  The  straight  line  joining  two  nonconsecutive 
vertices  of  any  figure  is  called  a  diagonal. 


60  BOOK  I.    PLANE  GEOMETRY 

Proposition  XXV.    Theorem: 

123.  TiDO  angles  whose  sides  are  parallel  each  to  each 
are  either  equal  or  supplementary. 


z 

Given  the  angle  AOB  and  the  lines  WY  and  XZ  parallel  to  the 
sides  and  intersecting  at  P,  the  figure  being  lettered  as  shown. 

To  prove  that  /.p  =  /LO,  and  that  Z.p'  is  supplementary 
to  ZO. 

Proof.    Let  OA  meet  XZ  at  M.    Then  in  the  figure 

ZO  =  Z 7/1,  and  Zp  =  Z m.  §  102 

{If  two  II  lines  are  cut  by  a  transversal,  the  ext.-int.  A  are  equal.) 

.■.Zp  =  Z.O.  Ax.  8 

Also  ZL])'  is  the  supplement  of  Zp.  §  42 

.*.  Zp'  is  supplementary  to  Z  0,  by  §  58.  q.e.d. 

If  the  sides  of  two  angles  are  parallel  each  to  each,  under  what 
circumstances  are  the  angles  equal,  and  under  what  circumstances  are 
they  supplementary  ? 

124.  Corollary.  The  opposite  angles  of  a  parallelogram 
are  equal,  and  any  two  consecutive  angles  are  supplementary. 

Draw  the  figure  and  explain  how  it  is  known  that  any  angle  is  the 
supplement  of  its  consecutive  angle.  If  two  opposite  angles  are  supple- 
ments of  the  same  angle,  show  that  §  58  applies. 


QUADRILATERALS  61 

Proposition  XXVL    Theorem 
125.  The  opposite  sides  of  a  i^arallelogram  are  equal. 


A 

Given  the  parallelogram  ABCD. 
To  prove  that     BC  =  AD,  and  AB  =  DC. 
Proof.  Draw  the  diagoDal  A  C. 

In  the  A  ABC  and  CDA, 

AC  =  AC,  Iden. 

ZBAC  ==  ZDCA, 

and  ZACB  =  Z  CA D.  §  100 

.-.  A  ABC  is  congruent  to  A  CDA.  §  72 

.-.  BC  =  AD,  and  AB  =  DC,  by  §  67.  O-e.d. 

126.  Corollary  1.  A  diagonal  divides  a  parallelogram  into 
two  congruent  triangles. 

Upon  what  theorem  does  this  depend  ? 

127.  Corollary  2.    Segments  of  parallel  lines  cut  off  hy 
parallel  lines  are  equal. 

How  does  this  follow  from  the  proposition  ? 

128.  Corollary  3.   Two parallellines  are  everywhere  equally 
distant  from  each  other.  A B 


\i  AB  and  CD  are  parallel,  what  can 

be  said  of  Js  dropped  from  any  points  in     

AB  to  CD  (§  127)  ?    Hence  what  may 

be  said  of  all  points  in  AB  with  respect  to  their  distance  from  CD  ? 


62  BOOK  L    PLANE  GEOMETRY 

Proposition  XXVII.    Theorem 

129.  If  the  opposite  sides  of  a  quadrilateral  are  equal, 
the  figure  is  a  parallelogram. 


Given  the  quadrilateral  ABCD^  having  BC  equal  to  AD^  and 
AB  equal  to  DC. 

To  prove  that  the  quadrilateral  ABCD  is  a  parallelogram. 

Proof.  Draw  the  diagonal  A  C. 

In  the  A  ABC  ^nd'CDA, 

BC  =  AD,  Given 

AB  =  DC,  Given 

and                                         AC  =  AC.  Iden. 

.-.  A  ABC  is  congruent  to  A  CDA.  §  80 

( Two  A  are  congruent  if  the  three  sides  of  the  one  are  equal  respectively 
to  the  three  sides  of  the  other.) 

.-.  ZBAC==ZDCA, 

and  ZACB  =  ZCAD.  §67 

.-.  AB  is  II  to  DC, 

and  BC  is  II  to  AD.  §  101 

{When  two  lines  in  the  same  plane  are  cut  by  a  transversal,  if  the 
alt. -int.  A  are  equal,  the  two  lines  are  II.) 

.-.the  quadrilateral  ABCD  is  a  O,  by  §  118.       Q.e.d. 
This  proposition  is  the  converse  of  Prop.  XXVI. 


QUADEILATERALS  63 

Proposition  XXVIII.    Theorem 

130.  If  two  sides  of  a  quadrilateral  are  equal  and 
parallel,  then  the  other  two  sides  are  equal  and  par- 
allel, and  the  figure  is  a  p)arallelogram. 


Given  the  quadrilateral  ABCD^  having  AB  equal  and  parallel 
to  DC. 

To  prove  that  the  quadrilateral  ABCD  is  a  parallelogram. 
Proof.  Draw  the  diagonal  A  C. 

In  the  A  ABC  and  CDA, 

AC  =  AC,  Iden. 

AB  =  DC,  Given 

and  ABAC  =  A  DC  A .  §  100 

{If  two  II  lines  are  cut  by  a  transversal,  the  alt.-int.  A  are  equal.) 

.-.A  ABC  is  congruent  to  A  CDA.  §  68 

{Two  A  are  congruent  if  two  sides  and  the  included  Z  of  the  one  are 
equal  respectively  to  two  sides  and  the  included  Z  of  the  other.) 

.\BC  =  AD, 

and  ZACB  =  Z  CAD.  §  67 

.-.  BC  is  II  to  AD.  §  101 

( When  two  lines  in  the  same  plane  are  cut  by  a  transversal,  if  the 
alt.-int.  A  are  equal,  the  two  lines  are  II.) 

But  AB  is  II  to  DC.  Given 

.-.  the  quadrilateral  ABCD  is  a  O,  by  §  118.       q.e.d. 


64  BOOK  I.    PLANE  GEOMETRY 

Proposition  XXIX.    Theorem 

131.  The  diagonals  of  a  parallelogram  bisect  each 
other. 

G 


A 

B 

Given  the 

parallelogram 

ABCD 

,  with  the 

diagonals 

AC  and  BD 

intersecting 

at  0. 

To  prove 

that 

A0  = 

OC, 

and 

B0  = 

01). 

Proof.  If  we  can  show  that  the  A  ABO  is  congruent  to  the 
AC  DO,  or  that  the  ABCO  is  congruent  to  the  ADAO,  the 
proposition  is  evidently  proved,  since  the  corresponding  sides 
of  the  congruent  triangles  will  be  equal. 

Now  in  the  A  ABO  and  CDO, 

AB  =  CD,  §125 

{The  opposite  sides  of  a  O  are  equal.) 

Z.BAO  =  Z.DCO, 

and  Z  OBA  =  Z  ODC.  §  100 

{If  two  parallel  lines  are  cut  by  a  transversal,  the  alternate-interior 
angles  are  equal.) 

.'.  AABOi^  congruent  to  A  CDO.  §  72 

{Two  A  are  congruent  if  two  A  and  the  included  side  of  the  one  are 
equal  respectively  to  two  A  and  the  included  side  of  the  other.) 

.\AO  =  OC, 

and  BO  =  OD.  §  67 

{Corresponding  parts  of  congruent  A  are  equal.)  Q.  E.  D. 


QUADRILATERALS  65 

Proposition  XXX.    Theorem 

132.  Two  j^ciraUelogranis  are  congruent  if  tivo  sides 
and  the  included  angle  of  the  one  are  equal  respectively 
to  tivo  sides  and  the  included  angle  of  the  other. 


A  B  A'  B' 

Given  the  parallelograms  ABCD  and  A'B'C^D',  with  AB  equal 
to  A'B',  AD  to  A'Z)',  and  angle  A  to  angle  A'. 

To  prove  that         the  UJ  are  congruent. 

Proof.    Place  the  EJABCD  upon  the  CJA'B'C'D'  so  that  AB 
shall  fall  upon  and  coincide  with  its  equal,  A'B'.  Post.  5 

Then  AD  will  fall  along  A'D\ 
(For  Z  A  is  given  equal  to  Z  A'.) 

and  D  will  fall  on  D'. 
{For  AD  is  given  equal  to  A^D^.) 

Now  DC  and  D'C  are  both  II  to  A  'B'  and  are  dl'awn  through  D'. 
.'.DC  will  fall  along  D'C. '  §  94 

(Through  a  given  point  only  one  line  can  be  drawn  W  to  a  given  line.) 

Also  BC  and  B'C'  are  both  II  to  A  'D'  and  are  drawn  through  B'. 

-  .'.  BC  will  fall  along  B'C.  §  94 

.-.  C  will  fall  on  C.  §  55 

.'.  the  two  [U  coincide  and  are  congruent,  by  §  66.    q.e.d. 

133.  Corollary.    Two  rectangles  having  equal  bases  and 

equal  altitudes  are  congruent. 

How  is  this  shown  to  be  a  special  case  under  the  above  proposition  ? 
What  sides  are  equal,  and  what  inchided  angles  are  equal  ? 


66  BOOK  I.    PLANE  GEOMETRY 

Proposition  XXXI.    Theorem 

134.  If  three  or  more  parallels  intercept  equal,  seg- 
ments  on  one  transversal,  they  intercept  equal  segme7its 
on  every  transversal. 


Given  the  parallels  AB^  CD^  EF,  GH,  intercepting  equal  segments 
JBD,  DF,  FH  on  the  transversal  BH^  and  intercepting  the  segments 
AC  J  CEy  EG  on  another  transversal. 

To  prove  that  AC  =  CE  =  EG. 

Proof.    Suppose  AP,  CQ,  and  ER  drawn  II  to  BH. 

A  APC,  CQE,  ERG  =  A  BDC,  DFE,  FHG  respectively.  §  102 

But  A  BDC,  DFE,  FHG  are  equal.  §  102 

.-.  A  APC,  CQE,  ERG  are  equal.  Ax.  8 

AP,  CQ,  ER  are  parallel.  §  96 

Also  A  CAP,  ECQ,  GER  are  equal.  §  102 

Now  AP  =  BD,  CQ  =  DF,  ER  =  FH.  §  127 
{Segments  of  parallels  cut  off  by  parallels  are  equal.) 

But                                  BD  =  DF=FH.  Given 

.'.  AP  =  CQ  =  ER.  Ax.  8 

.'.  A  CPA,  EQC,  and  GRE  are  congruent.  §  72 

.'.AC  =  CE  =  EG,  by  §  67.  Q.e.d. 


QUADRILATERALS 


67 


135.  Corollary  1.  If  a  line  is  parallel  to  one  side  of  a  tri- 
angle and  bisects  another  side,  it  bisects  the  third  side  also. 

Let  BE  be  II  to  BC  and  bisect  AB.  Suppose  a  line  is  drawn  through 
A  II  to  BC.  Then  how  do  we  know  this  line  to 
be  II  to  DE  ?  Since  it  is  given  that  the  three 
lis  intercept  equal  segments  on  the  transversal 
AB,  what  can  we  say  of  the  intercepted  seg- 
ments on  AC  ?  What  can  we  then  say  that  DE 
does  to  AC? 

Write  the  proof  of  this  corollary  in  full. 

136.  Corollary  2.  The  line  ivhich  joins  the  mid-points 
of  two  sides  of  a  triangle  is  parallel  to  the  third  side,  and  is 
equal  to  half  the  third  side. 

A  line  DE  drawn  through  the  mid-point  of  AB,  II  to  BC,  divides  AC 
in  what  way  (§  135)  ?  Therefore  the  line  joining  the  mid-points  oi  AB 
and  AC  coincides  with  this  parallel  and  is  II  to 
BC.  Also  since  JPF drawn  II  to  AB  bisects  AC, 
how  does  it  divide  BC  ?  What  does  this  prove 
as  to  the  relation  of  BF,  FC,  and  BC  ?  Since 
BFED  is  a  O  (§118),  what  do  we  know  as  to 
the  equality  of  DE,  BF,  and  i  BC  ? 

Write  the  proof  of  this  corollaiy  in  full. 

137.  Corollary  3.  Tlie  line  joining  the  mid-points  of  the 
nonparallel  sides  of  a  trapezoid  is  parallel  to  the  bases  and 
is  equal  to  half  the  sum  of  the  bases. 


F^-^ 


A 


B 


Draw  the  diagonal  DB.  In  the  A  ABD 
join  E,  the  mid-point  of  AD,  to  F,  the  mid- 
point of  DB.  Then,  by  §  136,  what  relations 
exist  between  EF  and  AB?  In  the  A  DBC 
join  F  to  G,  the  mid-point  of  BC.  Then  what 
relations  exist  between  FG  and  DC?   Since 

this  relation  exists,  what  relation  exists  between  AB  and  FG  ?  But  only 
one  line  can  be  drawn  through  F  W  to  AB  {%  94).  Therefore  FG  is  the 
prolongation  of  EF.  Hence  EFG  is  parallel  to  AB  and  CD,  and  equal 
to  ^{AB-k- DC). 

Write  the  proof  of  this  corollary  in  full. 


68  BOOK  I.    PLAXE  GEOMETRY 

138.  Polygon.    A  portion  of  a  plane  bounded  by  a  broken 
line  is  called  ?^  polygon. 

The  terms  sides,  perimeter,  angles,  vertices,  and  diagonals  are  employed 
in  the  usual  sense  in  connection  with  polygons  in  general. 

139.  Polygons  classified  as  to  Sides.    A  polygon  is 

a  triangle,  if  it  has  three  sides ; 
a  quadrilateral,  if  it  has  four  sides ; 
a  pentagon,  if  it  has  five  sides ; 
a  hexagon,  if  it  has  six  sides. 

These  names  are  sufficient  for  most  cases.    The  next  few  names  in 
order  are  heptagon,  octagon,  nonagon,  decagon,  undecagon,  dodecagon. 

A  polygon  is  equilateral,  if  all  of  its  sides  are  equal. 

140.  Polygons  classified  as  to  Angles.    A  polygon  is 
eqtdangular,  if  all  of  its  angles  are  equal ; 

convex,  if  each  of  its  angles  is  less  than  a  straight  angle ; 
concave,  if  it  has  an  angle  greater  than  a  straight  angle. 


Equilateral  Equiangular  Hexagon  Convex       Concave 

An  angle  of  a  polygon  greater  than  a  straight  angle  is  called  a  reentrant 
angle.    When  the  term  polygon  is  used,  a  convex  polygon  is  understood. 

141.  Regular  Polygon.  A  polygon  that  is  both  equiangular 
and  equilateral  is  called  a  regular  polygon. 

142.  Relation  of  Two  Polygons.    Two  polygons  are 
mutually  equiangular,  if  the  angles  of  the  one  are  equal  to 

the  angles  of  the  other  respectively,  taken  in  the  same  order ; 

TRutually  equilateral,  if  the  sides  of  the  one  are  equal  to  the 
sides  of  the  other  respectively,  taken  in  the  same  order ; 

congruent,  if  mutually  equiangular  and  mutually  equilateral, 
since  they  then  can  be  made  to  coincide. 


POLYGONS 


69 


Proposition  XXXII.    Theorem 

143.  The  sum  of  the  interior  angles  of  a  polygon  is 
equal  to  two  right  angles,  taken  as  r)iamj  times  less  two 
as  the  figure  has  sides. 


Given  the  polygon  ABCDEFy  having  n  sides. 

To  ])rove  that  the  sum  of  the  interior  A  =(n—  2^2  rt,  A. 

Proof.    From  A  draw  the  diagonals  AC,  AD,  AE. 

The  sum  of  the  A  of  the  A  is  equal  to  the  sum  of  the  A  of 
the  polygon.  Ax.  11 

Now  there  are  (n  —  2)  A. 
{For  there  is  one  A  for  each  side  except  the  two  sides  adjacent  to  A.) 

The  sum  of  the  A  of  each  A  =  2  rt.  ^.  §  107 

.*.  the  sum  of  the  A  of  the  (n  —  2)  A,  that  is,  the  sum  of  the 

A  of  the  polygon,  is  equal  to  (n  —  2)2  rt.  A,  by  Ax.  3.      q.e.d. 

144.  Corollary  1.  The  sum  of  the  angles  of  a  quadrilateral 
equals  four  right  angles  ;  and  if  the  angles  are  all  equal,  each 
is  a  right  angle. 

145.  Corollary  2.  Uach  angle  of  a  regular  polygon  of  n 
sides  is  equal  to  — '-  right  angles. 


n 


70  BOOK  I.    PLANE  GEOMETRY 

EXERCISE  12 

1.  What  is  the  sum  of  the  angles  of  («)  a  pentagon  ?  (h)  a 
hexagon  ?  (c)  a  heptagon  ?  (d)  an  octagon  ?  (6)  a  decagon  ? 
(/)  a  dodecagon?   (^)  a  polygon  of  24  sides? 

2.  What  is  the  size  of  each  angle  of  (<()  a  regular  pentagon  ? 
(^)  a  regular  hexagon  ?  (c)  a  regular  octagon  ?  (c?)  a  regular 
decagon  ?   (e)  a  regular  polygon  of  32  sides  ? 

3.  How  many  sides  has  a  regular  polygon,  each  angle  of 
which  is  1|  right  angles  ? 

4.  How  many  sides  has  a  regular  polygon,  each  angle  of 
which  is  If  right  angles  ? 

5.  How  many  sides  has  a  regular  polygon,  each  angle  of 
which  is  108°? 

6.  How  many  sides  has  a  regular  polygon,  each  angle  of 
which  is  140°? 

7.  How  many  sides  has  a  regular  polygon,  each  angle  of 
which  is  156°? 

8.  Four  of  the  angles  of  a  pentagon  are  120°,  80°,  90°,  and 
100°  respectively.    Find  the  fifth  angle. 

9.  Five  of  the  angles  of  a  hexagon  are  100°,  120°,  130°,  150°, 
and  90°  respectively.    Find  the  sixth  angle. 

10.  The  angles  of  a  quadrilateral  are  x,  2x,  2x,  and  3x. 
How  many  degrees  are  there  in  each  ? 

11.  The  angles  of  a  quadrilateral  are  so  related  that  the  sec- 
ond is  twice  the  first,  the  third  three  times  the  first,  and  the 
fourth  four  times  the  first.    How  many  degrees  in  each  ? 

12.  The  angles  of  a  hexagon  are  x,  3x,  3x,  2x,  2x,  and  x. 
How  many  degrees  are  there  in  each  ? 

13.  The  sum  of  two  angles  of  a  triangle  is  100°  and  their 
difference  is  40°.  How  many  degrees  are  there  in  each  of  the 
three  angles  of  the  triangle  ? 


POLYGONS  71 

Proposttton  XXXTir.    Theorem 

146.  Tlie  sum  of  the  exterior  angles  of  a  polijcjon, 
made  by  j^'^oducing  each  of  its  sides  hi  succession,  is 
equal  to  four  right  angles. 


Given  the  polygon  ABCDE,   having   its   n  sides    produced  in 
succession. 

To  prove  that  the  sum  of  the  exterior  A  =  4:rt.  A. 

Proof.    Denote  the  interior  A  of  the  polygon  by  a,  b,  c,  d,  e, 
and  the  corresponding  exterior  A  by  a',  b',  c',  d',  e'. 
Then,  considering  each  pair  of  adjacent  angles, 

A  a  -j-  Aa'=  a  st.  A, 

and  Ab-{-Ab'=  ^st  A.  §43 

( The  two  adjacent  A  which  one  straight  line  makes  with  another  are 
together  equal  to  a  straight  Z.) 

In  like  manner,  each  pair  of  adj.  zi  =  a  st.  A. 
But  the  polygon  has  n  sides  and  n  angles. 
Therefore  the  sum  of  the  interior  and  exterior  A§  is  equal 
to  n  st.  A,  OT  2n  rt.  A.  Ax.  3 

But  the  sum  of  the  interior  A  =  (n-2)2  rt.  A  §  143 

=  2n  rt.  Zs  -  4  rt.  A. 
.*.  the  sum  of  the  exterior  A  =  4:  rt.  A,  by  Ax.  2.     q.e.d. 


72  BOOK  I.    PLANE  GEOMETRY 

EXERCISE  13 

1.  An  exterior  angle  of  a  triangle  is  130°  and  one  of  the 
opposite  interior  angles  is  52°.  Find  the  number  of  degrees  in 
each  angle  of  the  triangle. 

2.  Two  consecutive  angles  of  a  rectangle  are  bisected  by 
lines  meeting  at  P.    How  many  degrees  in  the  angle  P  ? 

3.  Two  angles  of  an  equilateral  triangle  are  bisected  by  lines 
meeting  at  P.    How  many  degrees  in  the  angle  P  ? 

4.  The  two  base  angles  of  an  isosceles  triangle  are  bisected 
by  lines  meeting  at  P.  The  vertical  angle  of  the  triangle  is  30°. 
How  many  degrees  in  the. angle  P? 

5.  The  vertical  angle  of  an  isosceles  triangle  is  40°.  This 
and  one  of  the  base  angles  are  bisected  by  lines  meeting  at  P. 
How  many  degrees  in  the  angle  P  ? 

6.  One  exterior  angle  of  a  parallelogram  is  one  eighth  of  the 
sum  of  the  four  exterior  angles.  How  many  degrees  in  each 
angle  of  the  parallelogram  ? 

7.  How  many  degrees  in  each  exterior  angle  of  a  regular 
hexagon  ?  of  a  regular  octagon  ? 

8.  In  a  right  triangle  one  acute  angle  is  twice  the  other. 
How  many  degrees  in  each  exterior  angle  of  the  triangle  ? 

9.  Make  out  a  table  showing  the  number  of  degrees  in  each 
interior  angle  and  each  exterior  angle  of  regular  polygons  of 
three,  four,  five,  •  •  • ,  ten  sides. 

10.  If  the  diagonals  of  a  quadrilateral  bisect  each  other,  the 
figure  is  a  parallelogram. 

11.  In  this  parallelogram  ABCD,  AP  = 
CR,  and  BQ  =  DS.    Prove  that  PQRS  is    ^ 
also  a  parallelogram.  A      F  b 

12.  If  the  mid-points  of  the  sides  of  a  parallelogram  are 
connected  in  order,  the  resulting  figure  is  also  a  parallelogram. 


LOCI  OF  POINTS  73 

147.  Locus.  The  path  of  a  point  that  moves  in  accordance 
with  certain  given  geometric  conditions  is  called  the  locus  of 
the  point. 

Thus,  considering  only  figures  in  a  plane,  a  . 
point  at  a  given  distance  from  a  given  line  of  ^ 
indefinite  length  is  evidently  in  one  of  tw^o  lines 
parallel  to  the  given  line  and  at  the  given  distance  from  it.  Thus,  if  ^jB 
is  the  given  line  and  d  the  given  distance,  the  locus  is  evidently  the 
pair  of  parallel  lines  XY  and  X'Y\  ^ ^ 

The  locus  of  a  point  in  a  plane  at  a  given  distance  r     /  \ 

from  a  given  point  O  is  evidently  the  circle  described  about    <  r_\ 

0  as  a  center  with  a  radius  r.  \        ^        I 

The  plural  of  locus  (a  Latin  v^^ord  meaning  "place")  is      \^         ^/ 
loci  (pronounced  lo-si). 

We  may  think  of  the  locus  as  the  place  of  all  points  that  satisfy  cer- 
tain given  geometric  conditions,  and  speak  of  the  locus  of  points.  Both 
expressions,  locus  of  a  point  and  locus  of  points,  are  used  in  mathematics. 


EXERCISE  14 

State  without  proof  the  following  loci  in  a  plane  : 

1.  The  locus  of  a  point  2  in.  from  a  fixed  point  O. 

2.  The  locus  of  the  tip  of  the  minute  hand  of  a  watch. 

3.  The  locus  of  the  center  of  the  hub  of  a  carriage  wheel 
moving  straight  ahead  on  a  level  road. 

4.  The  locus  of  a  point  1  in,  from  each  of  two  parallel  lines 
that  are  2  in.  apart. 

5.  The  locus  of  a  point  on  this  page  and  1  in.  from  the  edge. 

6.  The  locus  of  the  point  of  a  round  lead  pencil  as  it  rolls 
along  a  desk. 

7.  The  locus  of  the  tips  of  a  pair  of  shears  as  they  open, 
provided  the  fulcrum  (bolt  or  screw)  remains  always  fixed  in 
one  position. 

8.  The  locus  of  the  center  of  a  circle  that  rolls  around  another 
circle,  always  just  touching  it. 


74  BOOK  I.    PLANE  GEOMETRY 

148.  Proof  of  a  Locus.  To  prove  that  a  certain  line  or  group 
of  lines  is  the  locus  of  a  point  that  fulhlls  a  given  condition,  it 
is  necessary  and  sufficient  to  prove  two  things : 

1.  That  any  point  In  the  sujyposed  locus  satisfies  the  condition. 

2.  That  any  point  outside  the  supposed  locus  does  not  satisfy 
the  given  condition. 

For  example,  if  we  wish  to  find  the  locus  of 
a  point  equidistant  from  these  intersecting  lines 
AB^  CD^  it  is  not  sufficient  to  prove  that  any 
point  on  the  angle-bisector  PQ  is  equidistant  from 
AB  and  CD^  because  this  may  be  only  part  of  the  locus.  It  is  necessary 
to  prove  that  no  point  outside  of  PQ  satisfies  the  condition.  In  fact,  in 
this  case  there  is  another  line  in  the  locus,  the  bisector  of  the  Z  BOD, 
as  will  be  shown  in  §  152. 

149.  Perpendicular  Bisector.  A  line  that  bisects  a  given  line 
and  is  perpendicular  to  it  is  called  the  per2Je7idicida7'  bisector  of 
the  line. 

EXERCISE  15 

Draw  the  following  loci,  giving  no  proofs  : 

1.  The  locus  of  a  point  \  in.  below  the  base  of  a  given 
triangle  ABC. 

2.  The  locus  of  a  point  ^  in.  from  a  given  line  AB. 

3.  The  locus  of  a  point  1  in.  from  a  given  point  O. 

4.  The  locus  of  a  point  ^  in.  outside  the  circle  described 
about  a  given  point  O  with  a  radius  1\  in. 

5.  The  locus  of  a  point  \  in.  within  the  circle  described 
about  a  given  point  0  with  a  radius  1^  in. 

6.  The  locus  of  a  point  ^  in.  from  the  circle  described  about 
a  given  point  0  with  a  radius  1^  in. 

7.  The  locus  of  a  point  ^  in.  from  each  of  two  given  parallel 
lines  that  are  1  in.  apart. 


LOCI  OF  POINTS  75 

Proposition  XXXIV.    Theorem 

150.  The  locus  of  a  point  equidistant  from  the  extrem- 
ities of  a  given  line  is  the  perpendicular  bisector  of 
that  line.  ^ 


b['\ 


Given  FO,  the  perpendicular  bisector  of  the  line  AB. 

To  prove  that  YO  is  the  locus  of  a  point  equidistant  from 
A  and  B. 

Proof.    Let  P  be  any  point  in  YO,  and  C  any  point  not  in  YO. 

Draw  the  lines  PA,  PB,  CA,  and  CB. 

Since           .                        AO  =  BO,  Given 

and                                         OP  =  OP,  Iden. 

.-.  rt.  A  yl  OP  is  congruent  to  rt.  A  BOP.  §  90 

.\PA=PB.  §67 
Let  CA  cut  the  X  at  D,  and  draw  DB. 
Then,  as  above,                 DA  =  DB. 

But.                                     CB<CD-{-DB.     -  '—  Post.  3 

.-.  CB<CD  +  DA.  Ax.  9 

.'.  CB<CA.  Ax.  11 

.*.  FO  is  the  required  locus,  by  §  148.  q.e.d. 

151.  Corollary.  Two  points  each  equidistant  from  the 
extremities  of  a  line  determine  the  perpendicular  bisector  of 
the  line. 


76 


BOOK  I.    PLANE  GEOMETRY 


Proposition  XXXV.    Theorem 

152.  The  locus  of  a  point  equidistant  from  tivo  given 
intersecting  lines  is  a  pair  of  lines  hisecting  the  angles 
formed  hy  those  lines. 


Given  XX^  and  YY^  intersecting  at  O,  -4C  the  bisector  of  angle 
X'OF,  and  BD  the  bisector  of  angle  YOX. 

To  prove  that  the  pair  of  lines  AC  and  BD  is  the  locus  of 
a  point  equidistant  from  XX'  and  YY', 

Proof.    Let  P  be  any  point  on  A  C  or  BD,  and  Q  any  point  not  on 
AC  or  BD.    Let  PM  and  QR  be  X  to  XX',  PN  and  QS  to  YY'. 

Since                            Z  MOP  =  Z  PON,  Given 

and                                           OP  =  OP,  Iden. 

.'.  rt.  A  OMP  is  congruent  to  rt.  A  ONP.  §  91 

.\PM=PN.  §67 

Let  QS  cut  AO  at  P'.    Draw  PT  J_  to  XX',  and  draw  QT. 
Then,  as  above,  P'T=  P'S. 

But  P'r-{-P'Q>QT,  Post.  3 

and  QT>QR.  §86 

.-.  P'T-\-P'Q>QR.  Ax.  10 

Substituting,  P'5  +  ^'Q  >  QR,  or  Q^^  >  Q/?.  Ax.  9 

.'.  the  pair  of  lines  is  the  required  locus,  by  §  148.     q.e.d. 


METHODS  OF  PKOOF  77 

153.  The  Synthetic  Method  of  Proof.  The  method  of  proof  in 
which  known  truths  are  put  together  in  order  to  obtain  a  new 
truth  is  called  the  synthetic  method. 

This  is  the  method  used  in  most  of  the  theorems  already  given.  The 
proposition  usually  suggests  some  known  propositions  already  proved, 
and  from  these  v^^e  proceed  to  the  proof  required.  The  exercises  on  this 
page  and  on- pages  78  and  79  may  be  proved  by  the  synthetic  method. 

154.  Concurrent  Lines.  If  two  or  more  lines  pass  through  the 
same  point,  they  are  called  concurrent  lines. 

155.  Median.  A  line  from  any  vertex  of  a  triangle  to  the 
mid-point  of  the  opposite  side  is  called  a  median  of  the  triangle. 

EXERCISE  16 

1.  If  two  triangles  have  two  sides  of  the  one  equal  respec- 
tively to  two  sides  of  the  other,  and  the  angles  opposite  two  equal 
sides  equal,  the  angles  opposite  the  other  two  equal  sides  are 
equal  or  supplementary,  and  if  equal  the  triangles  are  congruent. 

Let  ^C  =  A'C\  BC  =  B'C\  and  /.B  =  /.B\ 

Place  AA'B'C  on  A  ABC  so  that  B'C  shall  coincide  with  BC,  and 
ZA^  and  ZA  shall  be  on  the  same  side  of  BC. 


Since  ZB'=  ZB,  B'A'  will  fall  along  what  line  ?  Then  A'  will  fall  at 
A  or  at  some  other  point  in  BA,  as  B.  If  A'  falls  at  A,  what  do  we  know 
about  the  congruency  of  the  AA'B'C  and  ABC  ? 

If  A^  falls  at  D,  what  about  the  congruency  of  the  A  A'B'C  and  BBC  ? 

Since  CB  =  C'A'  =  CA,  what  about  the  relation  of  Z  ^  to  Z  CBA  ? 

Then  what  about  the  relation  of  the  A  CBA  and  BBC  ? 

Then  what  about  the  relation  of  the  A  A  and  BBC  ? 

Draw  figures  and  show  that  the  triangles  are  congruent : 

1.  If  the  given  angles  B  and  JB'  are  both  right  or  both  obtuse. 

2.  If  the  angles  A  and  A^  are  both  acute,  both  right,  or  both  obtuse* 

3.  If  ^C  and  A'C  are  not  less  than  BC  and  B'C  respectively. 


78 


BOOK  I.    PLANE  CxEOMETEY 


2.  The  bisectors  of  the  angles  of  a  triangle  are  concurrent  in 
a  point  equidistant  from  the  sides  of  the  triangle. 

The   bisectors   of   two   angles,    as  AD   and   BE,  intersect   as   at  0. 
Why?    Now  show  that  0  is  equidistant  from  AC  and  ^ 

AB,  also  from  BC  and  AB,  and  hence  from  AC  and 
BC.   Therefore,  where  does  0  lie  with  respect  to  the         ,_g,,  p 
bisector  Ci^?  ^^l^^ 

This  point  0  is  called  the  incenter  of  the  triangle.  ^  p 

3.  The  perpendicular  bisectors  of  the  sides  of  a  triangle  are 
concurrent  in  a  point  equidistant  from  the  vertices. 

The  ±  bisectors  of  two  sides,  as  QQ^  and  RW,  intersect  as  at  0. 
Why  ?    Now  show  that  0  is  equidistant  from  B 
and  O,  also  from  C  and  A,  and  hence  from  A 
and  B.   Therefore,  where  does  0  lie  with  respect 
to  the  ±  bisector  PP'  ? 

This  point  0  is  called  the  circumcenter  of  the 
triangle. 

4.  The  perpendiculars  from  the  vertices  of  a  triangle  to  the 
opposite  sides  are  concurrent. 

Let  the  Js  be  J.Q,  BR,  and  CP.   Through  A,  B,  C  suppose  B'C%  C'A\ 


Bx 


-lA' 


& 


and  ^'i^'  drawn  II  to  CB,  AC,  and  BA  respec- 
tively. Now  show  that  C" J.  ==50  =  ^B'.  In  the 
same  way,  what  are  the  mid-points  of  C'A'  and 
A'B'  ?  How  does  this  prove  that  A  Q,  BR,  and  CP  J 

are  the  _L  bisectors  of  the  sides  of  the  A  A''B'C'  ? 
Proceed  as  in  Ex.  3. 

This  point  0  is  called  the  orthocenter  of  the  triangle: 

5.  The  medians  of  a  triangle  are  concurrent  in  a  point  two 
thirds  of  the  distance  from  each  vertex  to  the  middle  of  the 
opposite  side. 

Two  medians,  as  ^  Q  and  CP,  meet  as  at  0.  If  Y  is  the  mid-point  of  A  0, 
and  X  of  CO,  show  that  YX  and  PQ  are  II  to  ^C  and 
equal  to  ^  AC.  Then  show  that  AY=YO=OQ,  and 
CX  =  XO  =  OP.  Hence  any  median  cuts  off  on  any 
other  median  what  part  of  the  distance  from  the  ver- 
tex to  the  mid-point  of  the  opposite  side  ? 

This  point  0  is  called  the  centroid  of  the  triangle. 


METHODS  OF  PROOF  79 

6.  The  bisectors  of  two  vertical  angles  are  in  b 

the  same  straight  line.  ol^^^ 

c — ^^  — -^ 

7.  The  bisector  of  one  of  two  vertical  angles   ^/</ 

bisects  the  other.  d 

8.  The  bisectors  of  two  supplementary  adjacent  angles  are 
perpendicular  to  each  other. 

9.  The  bisectors  of  the  two  pairs  of  vertical  angles  formed 
by  two  intersecting  lines  are  perpendicular  to  each  other. 

10.  If  the  bisectors  of  two  adjacent  angles  are        JY        .b 


perpendicular  to  each  other,  the  adjacent  angles  \/^^ 

are  supplementary.  o 

11.  If  an  angle  is  bisected,  and  if  a  line  is  drawn      ^   ^^^    ^ 
through  the  vertex  perpendicular  to  the  bisector, 
this  line  forms  equal  angles  with  the  sides  of  the 
given  angle.  o       ^ 

12.  The  bisector  of  the  vertical  angle  of  an  isosceles  triangle 
bisects  the  base  and  is  perpendicular  to  the  base. 

13.  The  perpendicular  bisector  of  the  base  of  an  isosceles 
triangle  passes  through  the  vertex  and  bisects  the  e 
angle  at  the  vertex. 

14.  If  the  perpendicular  bisector  of  the  base  of 
a  triangle  passes  through  the  vertex,  the  triangle 
is  isosceles.  a       d      b 

15.  Any  point  in  the  bisector  of  the  vertical  angle  of  an  isos- 
celes triangle  is  equidistant  from  the  extremities  of  the  base. 

16.  If  two  isosceles  triangles  are  on  the  same  base,  a  line 
passing  through  their  vertices  is  perpendicular  to  the  base 
and  bisects  the  base. 

17.  Two  angles  whose  sides  are  perpendicular  each  to  each 
are  either  equal  or  supplementary. 

Under  what  circumstances  are  the  angles  equal,  and  under  what 
circumstances  are  they  supplementary  ? 


80  BOOK  I.    PLANE  GEOMETKY 

156.  The  Analytic  Method  of  Proof.  The  method  of  proof  that 
asserts  that  a  proposition  under  consideration  is  true  if  another 
proposition  is  true,  and  so  on,  step  by  step,  until  a  known 
truth  is  reached,  is  called  the  analytic  method. 

This  is  the  method  resorted  to  when  we  do  not  see  how  to  start  the 
ordinary  synthetic  proof.  The  exercises  on  this  page  and  on  pages  81 
and  82  may  be  investigated  by  the  analytic  method. 

EXERCISE  17 

1.  The  mid-point  of  the  hypotenuse  of  a  right  triangle  is 
equidistant  from  the  three  vertices. 

Given  M  the  mid-point  of  AC,  the  hypotenuse  of  the  rt.  /\ABC. 

To  prove  that  M  is  equidistant  from  A,  B,  and  C. 

We  may  reason  thus :  M  is  equidistant  from  A,  B,  and  C  if  AM=  BM. 
Why  is  this  the  case  ? 

AM=  BM  if  the  J.  MN  cuts  A  ABM  into  two 
congruent  A.  ]\] 

A  ANM  is  congruent  to  A  BNM  ii  AN=  NB. 

But  AN  does  equal  NB  (§  135),  because  MN 
is  II  to  CB,  and  AM  =  MC. 

Therefore  the  proposition  is  true. 

We  may  now,  in  writing  our  proof,  begin  with  this  last  step  and  work 
backwards,  as  in  the  synthetic  proofs  already  considered. 

2.  If  one  acute  angle  of  a  right  triangle  is  double  the  other, 
the  hypotenuse  is  double  the  shorter  side. 

Given  ZA  =  Z  a,  and  ZC  =  Z2a,  to  prove  that  ^C  is  double  BC. 
Let  M  be  the  mid-point  of  AC.  Then  ^C  is  double  BC  if  AM=BC. 
Why  ?  Now  if  we  draw  MN  II  to  CB,  what  can 
be  said  of  the  relation  of  AN  and  NB  ?  Why  ? 
Then  what  may  be  said  of  A  ANM  and  BNM? 
Why  ?  Then  what  may  be  said  of  AM  &nd  BM  ? 
of  Za  and  Zq?  Therefore  the  proposition  is 
true  if  BM  =  BC.  But  BM=  BC  ii  Z2a  =  Zr, 
or  if  Z2  a  =  Za-{-  Zq,  or  it  Za  =  Zq.  But  Za  =  Zq  because  we  have 
proved  that  AM  =  BM. 

Now  reverse  this  reasoning  and  write  the  proof  in  the  usual  synthetic 
form. 


METHODS  OF  PROOF  81 

3.  A  median  of  a  triangle  is  less  than  half  the  sum  of  the 
two  adjacent  sides. 

Given  CM  a  median  of  the  A  ABC. 

a 

To  prove  that        CM<i{BC  -\-  CA). 
Now  CM<i{BC  -\-CA), 

if  2CM<BC  +  CA.  "\       /J/ 

This  suggests  producing  CM  by  its  own  length  to  P,         \  / 
and  drawing  AP.  P 

Then  CP  =  2  CM, 

and  2 CM<BC  +  CA  if  CP<BC  -\-CA. 

But  CP<^P  +  Cy1.  Post.  3 

.-.  CP<BC-\-  CA  if  BC  =  AP, 

and  BC  =  AP  if  A  JlfBC  is  congraent  to  A  MAP.  §  67 

But  A  MBC  is  congraent  to  A  MAP,  §  68 

for  MB  =  MA,  Given 

CM=MP,  Hyp. 

and  ^JBJfC  =  Z^3fP.        \  §60 

.■.CP<BC-{-CA. 
.'.CM<i{BC  +  CA). 

4.  The  line  which  bisects  two  sides  of  a  ti'iangle  is  parallel 
to  the  third  side.  ^ 

Given  AD  equal  to  DB,  and  AE  equal  to  EC. 
To  prove  that  DE  is  II  to  BC.  Df 

Suppose  a  Une  drawn  from  C  II  to  BA,  and  suppose  DE 
produced  to  meet  it  at  G. 

DE  is  II  to  BC  if  BCGD  is  a  O. 

BCGD  is  a  O  if  CG  =  BD. 
CG  =  BD  if  each  is  equal  to  AD. 
Now  BD  =  AD, 

and  CG  =  ^D  if  A  CGE  is  congraent  to  A  ADE. 

But  A  CG^-E  is  congraent  to  A  ADE, 

for  EC  =  AE, 

Z.  CEG  =  Z  AED, 
and  ZGCE  =  ZA. 


82  BOOK  I.    PLANE  GEOMETRY 

5.  Two  isosceles  triangles  are  congruent  if  a  side  and  an 
angle  of  the  one  are  equal  respectively  to  the  corresponding 
side  and  angle  of  the  other. 

The  A  are  congruent  if  what  three  corresponding  parts  are  equal  ? 

6.  The  bisector  of  an  exterior  angle  of  an  isosceles  tri- 
angle,  formed  by  producing  one   of  the   equal 
sides  through  the  vertex,  is  parallel  to  the  base. 

AE  is  II  to  BC  if  what  angles  are  equal  ?    These  angles 
are  equal  if  Z  CAD  is  twice  what  angle  in  the  A  ?  ^ ^^ 

7.  If  one  of  the  equal  sides  of  an  isosceles  triangle  is  pro- 
duced through  the  vertex  by  its  own  length,  the  line  joining 
the  end  of  the  side  produced  to  the  nearer  end  of 
the  base  is  perpendicular  to  the  base. 

Z  DBA  is  a  rt.  Z  if  it  equals  the  sum  of  what  A  of  A  ABD  ?         ^ 
It  equals  this  sum  if  Zp  equals  what  angle  and  Z  q  equals 
what  other  angle  ? 

8.  If  the  equal  sides  of  an  isosceles  triangle  are  produced 
through  the  vertex  so  that  the  external  segments  are  equal, 
the  extremities  of  these  segments  are  equidistant  from  the 
extremities  of  the  base  respectively. 

9.  If  the  line  drawn  from  the  vertex  of  a  triangle  to  the 
mid-point  of  the  base  is  equal  to  half  the  base,  the  angle  at 
the  vertex  is  a  right  angle. 

10.  If  through  any  point  in  the  bisector  of  an  /      ^^^ 

angle  a  line  is  drawn  parallel  to  either  side  of 


the  angle,  the  triangle  thus  formed  is  isosceles,     o 

11.  Through  any  point  C  in  the  line  AJ3  an  intersecting  line 
is  drawn,  and  from  any  two  points  in  this  line  equidistant  from 
C  perpendiculars  are  drawn  to  AB  ov  AB  produced.  Prove  that 
these  perpendiculars  are  equal. 

12.  The  lines  joining  the  mid-points  of  the  sides 
of  a  triangle  divide  the  triangle  into  four  congruent     ^"^    ^"^ 
triangles. 


METHODS  OF  PEOOF  83 

157.  The  Indirect  Method  of  Proof.  The  method  of  proof  that 
assumes  the  proposition  false  and  then  shows  that  this  assump- 
tion is  absurd  is  called  the  indirect  method  or  the  reductio  ad 
ahsurdum. 

This  method  forms  a  kind  of  last  resort  in  the  proof  of  a  proposition, 
after  the  synthetic  and  analytic  methods  have  failed. 

EXERCISE  18 

1.  Given  ABC  and  ABD,  two  triangles  on  the  same  base  AB, 
and  on  the  same  side  of  it,  the  vertex  of  each  triangle  being 
outside  the  other  triangle.  Prove  that  ii  AC  equals 
AD,  then  BC  cannot  equal  BD. 

Assume  that  BC  =  BD  and  show  that  the  result  is  absurd, 
since  it  would  make  D  fall  on  C,  which  is  contrary  to  the 
given  conditions. 

2.  On  the  sides  of  the  angle  XOY  two  equal  segments  OA 
and  OB  are  taken.  On  AB  Si  triangle  APB  is  constructed  with 
AP  greater  than  J5P.  Prove  that  OP  cannot  y 
bisect  the  angle  XOY.                                            ^  <r^'^^'^^ 

Assume  that  OP  does  bisect  ZXOY.    What  is         ^^^^J/"^ 
the  result  ?    Is  this  result  possible  ?  ^ 

3.  From  M,  the  mid-point  of  a  line  AB,  MC  is  drawn  oblique 
to  AB.    Prove  that  CA  cannot  equal  CB.  c 

Assume  that  CA  does   equal  CB.    What  is  the 
result  ?    Is  this  result  possible  ? 

4.  If  perpendiculars  are  drawn  to  the  sides  ^  m  ^ 
of  an  acute  angle  from  a  point  within  the  angle,  they  cannot 
inclose  a  right  angle  or  an  acute  angle. 

Assume  that  they  inclose  a  right  angle  and  show  that  this  leads  to  an 
absurdity.    Similarly  for  an  acute  angle. 

5.  One  of  the  equal  angles  of  an  isosceles  triangle  is  five 
ninths  of  a  right  angle.  Prove  that  the  angle  at  the  vertex 
cannot  be  a  right  angle.  - 

Assume  that  it  is  a  right  angle.    Is  the  result  possible  ? 


84  BOOK  I.    PLANE  GEOMETKY 

158.  General  Suggestions  for  proving  Theorems.  The  following 
general  suggestions  will  often  be  helpful : 

1.  Draw  the  figures  as  accurately  as  possible. 

This  is  especially  helpful  at  first.  A  proof  is  often  rendered  difficult 
simply  because  the  figure  is  carelessly  drawn.  If  one  line  is  to  be  laid  off 
equal  to  another,  or  if  one  angle  is  to  be  made  equal  to  another,  do  this 
by  the  help  of  the  compasses  or  by  measuring  with  a  ruler. 

2.  Draw  as  general  figures  as  possible. 

If  you  wish  to  prove  a  proposition  about  a  triangle,  take  a  scalene  tri- 
angle. If  an  equilateral  triangle,  for  example,  is  taken,  it  may  lead  to 
believing  something  true  for  every  kind  of  a  triangle,  when,  in  fact,  it 
is  true  for  only  that  particular  kind. 

3.  After  drawing  the  figure  state  very  clearly  exactly  what 
is  given  and  exactly  tvhat  is  to  be  proved. 

Many  of  the  difficulties  of  geometry  come  from  failing  to  keep  in  mind 
exactly  what  is  given  and  exactly  what  is  to  be  proved. 

4.  Then  proceed  synthetically  with  the  proof  if  you  see  how 
to  begin.  If  you  do  not  see  how  to  begin,  try  the  analytic  method, 
stating  clearly  that  you  could  prove  this  if  you  could  prove  that, 
and  so  on  until  you  reach  a  known  proposition. 

5.  If  two  lines  are  to  be  proved  equal,  try  to  prove  them  corre- 
sponding sides  of  congruent  triangles,  or  sides  of  an  isosceles 
triangle,  or  opposite  sides  of  a  parallelogram,  or  segments  between 
parallels  that  cut  equal  segments  from  another  transversal. 

6.  If  two  angles  are  to  be  proved  equal,  try  to  prove  them 
alternate- interior  or  exterior-interior  angles  of  parallel  lines,  or 
corresponding  angles  of  congruent  triangles,  or  base  angles  of 
an  isosceles  triangle,  or  opposite  angles  of  a  parallelogram. 

7.  If  one  angle  is  to  be  j^^oved  greater  than  another,  it  is  prob- 
ably an  exterior  angle  of  a  triangle,  or  an  angle  opposite  the 
greater  side  of  a  triangle. 

8.  If  one  line  is  to  be  proved  greater  than  another,  it  is  prob- 
ably opposite  the  greater  angle  of  a  triangle. 


EXERCISES 


85 


EXERCISE  19 

Prove  the  following  propositions  referring  to  equal  lines  : 

1.  If  the  sides  AB  and  AD  oi  ?,  quad- 
rilateral ABCD  are  equal,  and  if  the  di- 
agonal AC  bisects  the  angle  at  A,  then 
EC  is  equal  to  DC. 

A  B 

2.  A  line  is  drawn  terminated  by  two  parallel  lines.  Through 
its  mid-point  any  line  is  drawn  terminated  by  the  parallels. 
Prove  that  the  second  line  is  bisected  by  the  first. 

3.  In  a  parallelogram  ABCD  the  line  BQ  

bisects  AD,  and  DP  bisects  BC.    Prove  that      Q, 
BQ  and  DP  trisect  AC. 


p   X 


4.  On  the  base  .45  of  a  triangle  ABC  any 
point  P  is  taken.  The  lines  AP,  PB,  BC,  and 
CA  are  bisected  by  W,  X,  Y,  and  Z  respec- 
tively.   Prove  that  X  F  is  equal  to  WZ.  ^       w 

5.  In  an  isosceles  triangle  the  medians  drawn  to  the  equal 
sides  are  equal. 

6.  In  the  square  ABCD,  CD  is  bisected  by  Q,  and  P  and  R 
are  taken  on  AB  so  that  AP  equals  BR.  Prove  that  PQ 
equals  RQ.  c 

7.  In  this  figure  AC  =  BC,  and ^P  =  BQ  = 
CR  =  CS.   Prove  that  QR  =  PS. 

8.  From  the  vertex  and  the  mid-points  of  the  equal  sides  of 
an  isosceles  triangle  lines  are  drawn  perpendicular  to  the  base. 
Prove  that  they  divide  the  base  into  four  equal  parts. 

9.  In  the  quadrilateral  ABCD  it  is  known 
that  AB  is  parallel  to  DC,  and  that  angle  C 
equals  angle  D.  On  CD  two  points  are  taken 
such  that  CP=DQ.    Prove  that  AP  =  BQ. 


D    Q 


PC 


c 


86  BOOK  I.    PLANE  GEOMETRY 

EXERCISE  20 

Prove  the  following  propositions  referring  to  equal  angles : 

1.  In  this  figure  it  is  given  that  AC  =  BC, 
and  that  BQ  and  AR  bisect  the  angles  YBC 
and  CAX  respectively.  Prove  that  AAPB 
is  isosceles. 

2.  If  through  the  vertices  of  an  isosceles 
triangle  lines  are  drawn  parallel  to  the  oppo- 
site sides,  they  form  an  isosceles  triangle. 

3.  If  the  vertical  angles  of  two  isosceles  triangles  coincide, 
the  bases  either  coincide  or  are  parallel. 

4.  In  which  direction  must  the  side  of  a 
triangle  be  produced  so  as  to  intersect  the 
bisector  of  the  opposite  exterior  angle  ? 

Consider  the  cases,  ZA<ZC,ZA  =  ZC,ZA>ZC. 

5.  The  bisectors  of  the  equal  angles  of  an  isosceles  triangle 
form,  together  with  the  base,  an  isosceles  triangle. 

6.  The  bisectors  of  the  base  angles  of  an  equilateral  triangle 
form  an  angle  equal  to  the  exterior  angle  at  the 
vertex  of  the  triangle. 

7.  If  the  bisector  of  an  exterior  angle  of  a 
triangle  is  parallel  to  the  opposite  side,  the  tri- 
angle is  isosceles. 

8.  A  line  drawn  parallel  to  the  base  of  an  isosceles  triangle 
makes  equal  angles  with  the  sides  or  the  sides  produced. 

9.  A  line  drawn  at  right  angles  to  AB,  the  base  of  an 
isosceles  triangle  ABC,  cuts  ^C  at  P  and  BC  produced  at  Q. 
Prove  that  PCQ  is  an  isosceles  triangle.  b d 

10.  In  this  figure,  it  AB  =  CD,  and  Z  ^  =  Z  C,     \  / 

then  BD  is  parallel  to  AC.  \ J, 


EXERCISES  87 

EXERCISE  21 

Prove  the  following  propositions  hy  showing  that  two  tri- 
angles are  congruent : 

1.  A  perpendicular  to  the  bisector  of  an  angle  forms  with 
the  sides  an  isosceles  triangle. 

2.  If  two  lines  bisect  each  other  at  right  angles,  any  point  in 
either  is  equidistant  from  the  extremities  of  the  other, 

3.  From  B  a  perpendicular  is  drawn  to  the  bisector  of  the 
angle  A  of  the  triangle  ABC,  meeting  it  at  X,  and  meeting  ^C 
or  AC  produced  at  Y.    Prove  that  BX  =  XY. 

4.  If  through  any  point  equally  distant  from  two  parallel 
lines  two  lines  are  drawn  cutting  the  parallels,  they  intercept 
equal  segments  on  these  parallels.  ^ 

5.  If  from  the  point  where  the  bisector  of  an 
angle  of  a  triangle  meets  the  opposite  side, 
parallels  are  drawn  to  the  other  two  sides,  and 
terminated  by  the  sides,  these  parallels  are  equal. 

6.  The  diagonals  of  a  square  are  perpendicular  to  each  other 
and  bisect  the  angles  of  the  square. 

7.  If  from  a  vertex  of  a  square  there  are  drawn  line-seg- 
ments to  the  mid-points  of  the  two  sides  not  adjacent  to  the 
vertex,  these  line-segments  are  equal. 

8.  If  either  diagonal  of  a  parallelogram  bisects  one  of  the 
angles,  the  sides   of   the  parallelogram  are  Q 

all  equal.  /   \^ 

9.  On  the  sides  of  any  triangle  ABC  equi-       /  \^ 

lateral  triangles  BPC,  CQA,  ARB  are  con-      /^.^--^/X 

structed.    Prove  that  AP  =  BQ  =  CR.  ^\~~^—/   ^P 

\  /'^ 

How  can  we  prove  that  A  ABP  is  congruent  to        \        / 

ARBC?  Also  that  A  ^fiC  is  congruent  to  A  ^^Q?  V/ 

Does  this  prove  the  proposition  ?  \B         "     "    -'' 


88  BOOK  I.    PLANE  GEOMETRY 

EXERCISE  22 

Prove  the  following  propositions  relating  to  the  sum  of  the 
angles  of  a  polygon  : 

1.  An  exterior  angle  of  an  acute  triangle  or  of  a  right 
triangle  cannot  be  acute. 

2.  If  the  sum  of  two  angles  of  a  triangle  equals  the  third 
angle,  the  triangle  is  a  right  triangle. 

3.  If  the  line  joining  any  vertex  of  a  triangle  to  the  mid- 
point of  the  opposite  side  divides  the  triangle  into  two  isos- 
celes triangles,  the  original  triangle  is  a  right  triangle. 

4.  If  the  vertical  angles  of  two  isosceles  triangles  are  sup- 
plements one  of  the  other,  the  base  angles  of  the  one  are 
complements  of  those  of  the  other. 

5.  From  the  extremities  of  the  base  AB  oi  Si 
triangle  ABC  perpendiculars  to  the  other  two  sides 
are  drawn,  meeting  at  P.    Prove  that  the  angle  P  is  ^ 
the  supplement  of  the  angle  C. 

6.  If  two  sides  of  a  quadrilateral  are  parallel,  and  the  other 
two  sides  are  equal  but  not  parallel,  the  sums  of  the  two  pairs 
of  opposite  angles  are  equal. 

7.  The  bisectors  of  two  consecutive  angles  of  a  parallelogram 
are  perpendicular  to  each  other. 

8.  The  exterior  angles  at  B  and  C  of  any 
triangle  ABC  are  bisected  by  lines  meeting 
at  P.  Prove  that  the  angle  at  P  together 
with  half  the  angle  A  equals  a  right  angle. 

9.  The  opposite  angles  of  the  quadrilateral  formed  by 
the  bisectors  of  the  interior  angles  of  any  quadrilateral  are 
supplemental. 

10.  Show  that  Ex.  9  is  true,  if  the  bisectors  of  the  exterior 
angles  are  taken. 


EXERCISES  89 

EXERCISE  23 

Prove  the  following  propositions  referring  to  greater  lines  or 
greater  angles : 

1.  In  the  triangle  ABC  the  angle  A  is  bisected  by  a  line 
meeting  BC  at  D.  Prove  that  BA  is  greater  than  BD,  and  CA 
greater  than  CD. 

2.  In  the  quadrilateral  ABCD  it  is  known  that  AD  is  the 
longest  side  and  BC  the  shortest  side.    Prove  that  the  angle  B 
is  greater  than  the  angle  D,  and  the  angle  C  greater  p 
than  the  angle  A.                                                              ^ 

3.  A  line  is  drawn  from  the  vertex  y4  of  a  square 
ABCD  so  as  to  cut  CD  and  to  meet  BC  produced 
in  P.    Prove  that  ^P  is  greater  than  DB. 

4.  If  the  angle  between  two  adjacent  sides  of  a  parallelo- 
gram is  increased,  the  length  of  the  sides  remaining  unchanged, 

the  diagonal  from  the  vertex  of  this  angle  is  diminished. 

c 

5.  Within  a  triangle  ABC  a  point  P  is  taken 

such  that  CP  =  CB.    Prove  that  AB  is  always 
greater  than  A  P. 

^  A  M     B 

6.  In  a  quadrilateral  ABCD  it  is  known  that  AD  equals  BC 
and  that  the  angle  C  is  less  than  the  angle  D.  Prove  that  the 
diagonal  ^C  is  greater  than  the  diagonal  BD. 

7.  In  the  quadrilateral  ABCD  it  is  known  that  AD  equals 
BC  and  that  the  angle  D  is  greater  than  the  angle  C.  Prove 
that  the  angle  B  is  greater  than  the  angle  .4,  c 

8.  In  the  triangle  ABC  the  side  AB  is  greater      ^/ 
than  A  C.  On  A  B  and  A  C  respectively  BP  is  taken 
equal  to  CQ.    Prove  that  Z>Q  is  greater  than  CP.   ^  ^ 

9.  The  sum  of  the  distances  of  any  point  from 
the  three  vertices  of  a  triangle  is  greater  than 
half  the  sum  of  the  sides. 


90 


BOOK  I.    PLANE  GEOMETRY 


EXERCISE  24 
Prove  the  following  miscellaneous  exercises  : 

1.  The  line  joining  the  mid-points  of  the  nonparallel  sides 
of  a  trapezoid  passes  through  the  mid-points  of      jy 
the  two  diagonals. 

How  is  EF  related  to  AB  and  BC  ?   Why  ? 
Since  EF  bisects  BC  and  AB,  how  does  it  divide  AC 
aiidBB?   Why? 

2.  The  lines  joining  the  mid-points  of  the 
consecutive  sides  of  any  quadrilateral  form 
a  parallelogram. 

How  are  PQ  and  SR  related  to  ^C  ?  ^ p b 

3.  If  the  diagonals  of  a  trapezoid  are  equal,  the  trapezoid  is 
isosceles.  c, d 

Draw  CE  and  BF  ±  to  AB. 

How  isAABF  related  to  A  BCE  ?   Why  ? 

Then  how  is  Z  FAB  related  to  Z  CBA  ? 

Then  how  \sAABC  related  to  A  BAB?   Why  ?  a    E  F    B 

4.  If  from  the  diagonal  DB,  of  a  square  ABCD,  BE  is  cut  off 
equal  to  5C,  and  EF  is  drawn  perpendicular  to 
BDj  meeting  DC  2X  F,  then  DE  is  equal  to  jBi^  and 
also  to  FC. 

How  many  degrees  in  A  EBF  and  BFE  ?    How  is  BE 
related  to J^F?   Why? 

Then  how  is  rt.  A  BEF  related  to  rt.  A  BCF  ?    Why  ? 

5.  If  the  opposite  sides  of  a  hexagon  are  equal  and  parallel, 
the  diagonals  that  join  opposite  vertices  meet  in  a  point. 

6.  If  perpendiculars  are  drawn  from  the  four 
vertices  of  a  parallelogram  to  any  line  outside  the 
parallelogram,  the  sum  of  the  perpendiculars  from 
one  pair  of  opposite  vertices  equals  the  sum  of 
those  from  the  other  pair. 

How  are  x  ■\-  y  and  to  -|-  2-  related  to  A:  ? 


z 

y    A 

k  /y/ 

EXERCISES  91 

EXERCISE  25 
Examination  Questions 

1.  The  sum  of  the  four  sides  of  any  quadrilateral  is  greater 
than  the  sum  of  the  diagonals. 

2.  The  lines  joining  the  mid-points  of  the  sides  of  a  square, 
taken  in  order,  form  a  square. 

3.  In  a  quadrilateral  the  angle  between  the  bisectors  of  two 
consecutive  angles  is  one  half  the  sum  of  the  other  two  angles. 

4.  If  the  opposite  sides  of  a  hexagon  are  equal,  does  it  follow 
that  they  are  parallel  ?    Give  reasons  for  your  answer. 

5.  In  a  triangle  ABC  the  side  BC  is  bisected  at  P  and  AB 
is  bisected  at  Q.  AP  is  produced  to  R  so  that  AP=  PR,  and 
CQ  is  produced  to  S  so  that  CQ  =  QS.  Prove  that  S,  B,  and  it 
are  in  a  straight  line. 

6.  If  the  diagonals  of  a  parallelogram  are  equal,  all  of  the 
angles  of  the  parallelogram  are  equal. 

7.  In  the  triangle  ABC,  ZA  =  60°  and  ZB>ZC.  Which 
is  the  longest  and  which  is  the  shortest  side  of  the  triangle  ? 
Prove  it. 

8.  How  many  sides  has  a  polygon  each  of  whose  interior 
angles  is  equal  to  175°  ? 

9.  Given  the  quadrilateral  ABCD,  with  AB  equal  to  AD, 
and  BC  equal  to  CD.  Prove  that  the  diagonal  A  C  bisects  the 
angle  DCB  and  is  perpendicular  to  the  diagonal  BD. 

10.  In  how  many  ways  can  two  congruent  triangles  be  put 
together  to  form  a  parallelogram  ?    Draw  the  diagrams. 

11.  The  sides  of  a  polygon  of  an  odd  number  of  sides  are 
produced  to  meet,  thus  forming  a  star-shaped  figure.  What  is 
the  sum  of  the  angles  at  the  points  of  the  star  ? 

The  propositions  in  Exercise  25  are  taken  from  recent  college  entrance 
examination  papers. 


92  BOOK  I.    PLANE  GEOMETRY 

EXERCISE  26 

Review  Questions 

1.  Define  and  illustrate  rectilinear  and  curvilinear  figures. 

2.  Upon  what  does  the  size  of  an  angle  depend  ? 

3.  What  is  meant  by  the  bisector  of  a  magnitude  ?    Illus- 
trate when  the  magnitude  is  a  line ;  an  angle. 

4.  Define  perpendicular  and  state  three  facts  relating  to  a 
perpendicular  to  a  line. 

5.  Name  and  define  the  parts  of  a  triangle  and  such  special 
lines  connected  with  a  triangle  as  you  have  thus  far  studied. 

6.  Classify  angles. 

7.  Classify  triangles  as  to  angles ;  as  to  sides. 

8.  Define  and  illustrate  complementary,  supplementary,  and 
conjugate  angles. 

9.  What  are  the  two  classes  of  assumptions  in  geometry  ? 
Give  the  list  of  each. 

10.  State  all  of  the  conditions  of  congruency  of  two  triangles. 

11.  What  is  meant  by  the  converse  of  a  proposition  ? 

12.  Are  two  triangles  always  congruent  if  three  parts  of  the 
one  are  respectively  equal  to  three  parts  of  the  other  ? 

13.  State  three  tests  for  determining  whether  one  line  is 
parallel  to  another. 

14.  State  the  proposition  relating  to  the  sum  of  the  angles  of 
a  triangle,  and  state  a  proposition  that  can  be  proved  by  its  use. 

15.  State  a  proposition  relating  to  two  unequal  angles  of  a 
triangle ;  to  two  unequal  sides  of  a  triangle. 

16.  Must  a  triangle  be  equiangular  if  equilateral  ?    Must  a 
triangle  be  equilateral  if  equiangular  ? 

17.  Classify  polygons  as  to  sides ;  as  to  angles. 

18.  Define  locus  and  give  three  illustrations. 


BOOK  II 

THE  CIRCLE 

159.  Circle.  A  closed  curve  lying  in  a  plane,  and  such  that 
all  of  its  points  are  equally  distant  from  a  fixed  point  in  the 
plane,  is  called  a  circle. 

160.  Circle  as  a  Locus.  It  follows  that  the  locus  of  a  point  in 
a  plane  at  a  given  distance  from  a  fixed  point  is  a  circle. 

161.  Radius.  A  straight  line  from  the  center  to  the  circle  is 
called  a  radius. 

162.  Equal  Radii.  It  follows  that  all  radii  of  the  same  circle 
or  of  equal  circles  are  equal,  and  that  all  circles  of  equal  radii 
are  equal. 

163.  Diameter.  A  straight  line  through  the  center,  termi- 
nated at  each  end  by  the  circle,  is  called  a  diameter. 

Since  a  diameter  equals  two  radii,  it  follows  that  all  diameters  of  the 
same  circle  or  of  equal  circles  are  equal. 

164.  Arc.    Any  portion  of  a  circle  is  called  an  arc. 

An  arc  that  is  half  of  a  circle  is  called  a  semicircle. 

An  arc  less  than  a  semicircle  is  called  a  minor  arc,  and  an  arc  greater 
than  a  semicircle  is  called  a  major  arc.  The  word  arc  taken  alone  is  gen- 
erally understood  to  mean  a  minor  arc. 

165.  Central  Angle.  If  the  vertex  of  an  angle  is  at  the  center 
of  a  circle  and  the  sides  are  radii  of  the  circle,  the  angle  is 
called  a  central  angle. 

An  angle  is  said  to  intercept  any  arc  cut  off  by  its  sides,  and 
the  arc  is  said  to  subtend  the  angle. 


94 


BOOK  II.    PLANE  GEOMETEY 
Proposition  I.    Theorem 


166.  Ill  the  same  circle  or  in  equal  ciixles  equal  cen- 
tral angles  intercept  equal  arcs  ;  and  of  tivo  unequal 
central  angles  the  greater  intercepts  the  greater  arc. 


Given  two  equal  circles  with  centers  0  and  0\  with  angles  AOB 
and  A'O'B'  equal,  and  with  angle  AOC  greater  than  angle  A'O'B^. 

To  prove  that       1.  arc  AB  =  are  A'B'; 

2.   arc  AC>  arc  A'B'. 

Proof.  1.  Place  the  circle  with  center  O  on  the  circle  with 
center  0'  so  that  Z  AOB  shall  coincide  with  its  equal,  AA'O'B'. 
In  the  case  of  the  same  circle,  swing  one  angle  about  0  until 
it  coincides  with  its  equal  angle. 

Then  A  falls  on  A',  and  B  on  B'. 
{Radii  of  equal  circles  are  equal.) 

.'.  arc  AB  coincides  with  arc  A'B'. 
{Every  point  of  each  is  equally  distant  from  the  center.) 

Proof.    2.  Since  Z^OC  is  greater  than  Z^'0'5', 
and  ZAOB  =  Z.A  'O'B', 

therefore  .    Z^OC  is  greater  than  Z. 4 (95. 

Therefore  OC  lies  outside  ZAOB. 

.'.  3iTG  AC>3i,TGAB. 

But  '  arc  ^5  =  arc  ^'^'. 

.*.  arc  ^  C>  arc  A  'B',  by  Ax.  9.  q.e.d. 


Post.  5 
§162 

§159 

Given 
Given 
Ax.  9 

Ax.  11 


CENTRAL  ANGLES  95 

Proposition  II.    Theorem 

167.  In  the  same  circle  or  in  equal  circles  equal  arcs 
subtend  equal  central  angles  ;  and  of  two  unequal  arcs 
the  greater  subtends  the  gi^eater  central  angle. 

Given  two  equal  circles  with  centers  0  and  O ',  with  arcs  AB  and 
A^B^  equal,  and  with  arc  AC  greater  than  arc  -4 '5'. 

To  prove  that       1.  AAOB  =  Z.  A'O'B'; 
2.  ZAOOZ. A'O'B'. 

Proof.   1.  Using  the  figure  of  Prop.  I,  place  the  circle  with 

center  O  on  the  circle  with  center  O'  so  that  OA  shall  fall  on 

its  equal  O'A',  and  the  arc  AB  on  its  equal  A'B'.  Post.  5 

Then  OB  coincides  with  O'B'.  Post.  1 

.■.ZAOB  =  Z  A'O'B'.  §23 

Proof.  2.  Since  arc  AC> arc  A'B',  it  is  greater  than  arc  AB, 

the  equal  of  arc  A'B',  and  OB  lies  within  the  Z.AOC.       Ax.  9 

.■.ZA0OZ.A0B.  Ax.  11 

.\ZA0OZ  A'O'B',  by  Ax.  9.         q.e.d. 

This  proposition  is  the  converse  of  Prop.  I. 

168.  Law  of  Converse  Theorems.  Of  four  magnitudes,  a,  h,  x,  y, 
if  (1)  a~>h  when  x  >  y, 

(2)  a  =  5  when  x  =  y, 

and  (3)  a<h  when  x<y, 

then  the  converses  of  these  three  statements  are  always  true. 

For  when  a  >  6  it  is  impossible  that  x  —  y^  for  then  a  would  equal  h 
by  (2) ;  or  that  x  <  y,  for  then  a  would  be  less  than  6  by  (3).  Hence  x  >  y 
when  a  >  6.    In  the  same  way,  x  =  2/  when  a  =  h,  and  x  <y  when  a  <b. 

169.  Chord.  A  straight  line  that  has  its  extremi-       ^■^^ss-*^ 
ties  on  a  circle  is  called  a  chord. 

A  chord  is  said  to  subtend  the  arcs  that  it  cuts  from  a 
circle.  Unless  the  contrary  is  stated,  the  chord  is  taken 
as  subtending:  the  minor  arc. 


96  BOOK  11.    PLANE  GEOMETRY 

Proposition  III.    Theorem 

170.  In  the  same  circle  or  in  equal  circles,  if  two 
arcs  are  equal,  they  are  subtended  hy  equal  chords ; 
and  if  ttvo  arcs  are  unequal,  the  greater  is  subtended 
by  the  greater  chord. 


Given  two  equal  circles  with  centers  O  and  0\  with  arcs  AB&nd 
A'B^  equal,  and  with  arc  AF  greater  than  arc  A^B\ 

To  prove  that    1.   chord  AB  =  chord  A' B' ; 
2.  chord  AF>  chord  A' B'. 

Proof.   1.  Draw  the  radii  OA,  OB,  OF,  O'A',  O'B'. 


Since 

OA  =  0'A',  and  OB  =  O'B', 

§162 

and 

ZAOB  =  ZA'0'B', 
{In  equal  d)  equal  arcs  siMend  equal  central  A.) 

§167 

.'.  A  0.45  is  congruent  to  A  O'A'B', 

§68 

and 

chord  .4  5  =  chord  .4 '5'. 

§67 

Proof. 

2.  In  the  A  OAF  and  O'A'B', 

OA  =  0'A',  and  OF  =  O'B', 

§162 

but 

Z  A  OF  is  greater  than  Z  A  'O'B'. 

§167 

(In  equal  (D,  of  two  unequal  arcs  the  greater  subtends  the  greater  central  Z. ) 
. • .  chord  AF>  chord  A 'B',  by  §  115.  Q. e. d. 

171.  Corollary.  In  the  same  circle  or  in  equal  circles,  the 
greater  of  two  unequal  major  arcs  is  subtended  hy  the  less  chord. 


ARCS  AND  CHORDS  97 

Proposition  IV.    Theorem 

172.  In  the  same  circle  or  in  equal  circles,  if  two 
chords  are  equal,  they  subtend  equal  arcs  ;  and  if  tivo 
chords  are  unequal,  the  greater  subtends  the  greater  arc. 


Given  two  equal  circles  with  centers  0  and  0\  with  chords  AB 
and  A'5'  equal,  and  with  chord  AF  greater  than  chord  A'B\ 

To  prove  that       1.  arc  AB  =  arc  A' B' ; 
2.  arcAF>arcA'B'. 

Proof.    1.  Draw  the  radii  OA,  OB,  OF,  O'A',  O'B'. 

Since                     OA  =  O'A',  and  OB  =  O'B',  §  162 

and                             chord  AB  =  chord  A  'B',  Given 

.'.AOABis  congruent  to  A  O'A'B',  §  80 

and                                  ZAOB  =  ZA  'O'B'.  §  67 

.-.  arc  AB  =  sltgA'B'.  §  166 

Proof.    2.  In  the  A  OAF  and  O'A'B', 

OA  =  O'A ',  and  OF  =  O'B',  §  162 

but  chord  ^F>  chord  ^'5'.  Given 

.\ZAOF>ZA'0'B'.  §116 

. • .  arc  ^  jP  >  arc  A  'B',  by  §  166.  Q. e.  d. 

This  proposition  is  the  converse  of  Prop.  III. 

173.  Corollary.  In  the  same  circle  or  in  equal  circles  the 
greater  of  two  unequal  chords  subtends  the  less  major  arc. 


98  BOOK  11.    PLANE  GEOMETRY 

Proposition  Y.    Theorem 

174.  A  line  through  the  center  of  a  circle  perpendicular 
to  a  chord  bisects  the  chord  and  the  arcs  subtended  by  it. 


Q 
Given  the  line  PQ  through  the  center  O  of  the  circle  AQBP^ 
perpendicular  to  the  chord  AB  at  M. 

To  prove  that  AM—  BM,  arc  AQ  =  arc  BQ,  and  arc  AP  = 
arc  BP. 

Proof.  Draw  the  radii  OA  and  OB. 

Then  since                       OM  =  OM,  Iden. 

and                                        OA  =  OB,  §  162 

.'.  rt.  A  A  MO  is  congruent  to  rt.  A  BMO.  §  89 

.\AM=BM,  3ind  ZA0Q  =  ZQ0B.  §67 

Likewise                    Z  POA  =  Z  BOP.  §  58 

. • .  arc  AQ  =  arc  BQ,  and  arc  AP  =  arc  BP,  by  §  166.  Q. e. d. 

175.  Corollary  1.  A  diameter  bisects  the  circle. 

176.  Corollary  2.   A  line  through  the  center  that  bisects 
a  chord  is  perpendicular  to  the  chord. 

177.  Corollary  3.    The  perpendicular  bisector  of  a  chord 

passes  through  the  center  of  the  circle  and  bisects  the  arcs 

subtended  by  the  chord. 

How  many  bisectors  of  the  chord  are  possible  ?    How  many  ±  bisec- 
tors ?   Therefore  with  what  line  must  this  coincide  (§  174)  ? 


ARCS  AND  CHORDS  99 

Proposition  VI.    Theorem 

178.  In  the  same  circle  or  in  equal  circles  equal  chords 
are  equidistant  from  the  center,  and  chords  equidistant 
from,  the  center  are  equal. 


Given  AB  and  CD,  equal  chords  of  the  circle  ACDB. 
To  prove  that  AB  and  CD  are  equidistant  from  the. center  0. 
Proof.         Draw  OP  ±toAB,  and  OQ  _L  to  CD. 
Draw  the  radii  OA  and  OC. 

OP  bisects  AB,  and  OQ  bisects  CD.  §  174 

Tlien  since                       AP  =  CQ,  Ax.  4 

and                                        OA  =  OC,  §  162 

.-.  rt.  A  OP  A  is  congruent  to  rt.  AOQC.  §  89 

.\OP  =  OQ.  §67 
.'.  AB  and  CD  are  equidistant  from  0,  by  §  88.     q.e.d. 

Given  OP  and  OQ^  equal  perpendiculars  from  the  center  O  to  the 
chords  AB  and  C2). 

2^0  prove  that  AB  =  CD. 

Proof.    Since                     OA  =  OC,  §  162 

and                                         OP  =  OQ,  Given 

.-.  rt.  A  OPA  is  congruent  to  rt.  A  OQC.  §  89 

.'.AP  =  CQ.  §67 

.-.  .4^  =  C/>,  by  Ax.  3.  Q.e.d. 


100  BOOK  II.    PLANE  GEOMETRY 

Proposition  VII.    Theorem 

179.  In  the  same  circle  or  in  equal  circles,  if  two 
chords  are  unequal,  they  are  unequally  distant  from 
the  center,  and  the  greater  chord  is  at  the  less  distance. 


Given  a  circle  with  center  0,  two  unequal  chords  AB  and  CD^ 
AB  being  the  greater,  and  OP  perpendicular  to  AB^  and  OQ  per- 
pendicular to  CD. 

To  prove  that  OP  <0Q. 

Proof.    Suppose  AE  drawn  equal  to  CD,  and  OP,  A-toAE. 
Draw  PR. 
OP  bisects  AB,  and  OR  bisects  AE.  §  174 

{A  line  through  the  center  of  a  circle  ±  to  a  chord  bisects  the  chord.) 

But  AB>  CD.  Given 

.-.  AB>AE,  the  equal  of  CD.  Ax.  9 

.-.  AP>AR.  Ax.  6 

.'.ZARP>ZRPA.  §113 

{If  two  sides  of  a  A  are  unequal,  the  A  opposite  these  sides  are  unequal, 
and  the  Z  opposite  the  greater  side  is  the  greater.) 

.'.  Z.PRO,  the  complement  of  ZARP,  is  less  than  Z.OPR, 
the  complement  of  Z  RPA.  §  59 

.'.OP<OR.  §114 

But  OR=OQ.  §178 

.-.  OP<OQ,  by  Ax.  9.  Q.e.d. 


AKCS  AND  CHORDS  101 

Proposition  VIII.    Theorem 

180.  In  the  same  circle  or  in  equal  circles,  if  tivo 
chords  are  unequally  distant  from  the  center,  they  are 
unequal,  and  the  chord  at  the  less  distance  is  the  greater. 


Given  a  circle  with  center  0,  two  chords  AB  and  CD  unequally 
distant  from  0,  and  OP^  the  perpendicular  to  AB^  less  than  OQ^ 
the  perpendicular  to  CD. 

To  prove  that  AB>CD. 

Proof.    Suppose  AE  drawn  equal  to  CD,  and  OR  X  to  AE. 

Now  OP  <  OQ,  Given 

and  OR  =  OQ.  §  178 

.-.  OP<OR.  Ax.  9 

Drawing  PR,  A  PRO  <  A  OPR.  §  1 1 3 

.•.  Z  ARP,  the  complement  of  Z  PRO,  is  greater  than  Z  RPA, 

the  complement  of  Z  OPR.  §  59 

.\AP>AR.  §114 

But  AP  =  \AB,2indiAR  =  \AE.  §174 

.'.AB>AE.  Ax.  6 

But  CD  =  AE.  Hyp. 

.*.  AB>CD,  by  Ax.  9.  Q.e.d. 
This  proposition  is  the  converse  of  Prop.  VII. 

181.  Corollary.  A  diameter  of  a  circle  is  greater  than 
any  other  chord. 


102 


BOOK  II.    PLANE  GEOMETRY 


182.  Secant.  A  straight  line  that  intersects  a  circle  is  called 
a  secant.    In  this  figure  AD  is  Si  secant. 

Since  only  two  equal  obliques  can  be  drawn 
to  a  line  from  an  external  point  (§  85),  and 
since  the  two  equal  angles  which  radii  make 
(§  74)  with  any  secant  where  it  cuts  the  circle 
cannot  be  right  angles  (§  109),  they  must  be 
oblique ;  and  hence  it  follows  that  a  secant  can 
intersect  the  circle  in  only  two  points. 

183.  Tangent.  A  straight  line  of  unlimited  length  that  has 
one  point,  and  only  one,  in  common  with  a  circle  is  called  a 
tangent  to  the  circle. 

In  this  case  the  circle  is  said  to  be  tangent  to  the  line.  Thus  in  the 
figure,  BC  is  tangent  to  the  circle,  and  the  circle  is  tangent  to  BC. 

The  common  point  is  called  the  point  of  contact  or  point  of  tangency. 

By  the  tangent  from  an  external  point  to  a  circle  is  meant  the  line- 
segment  from  the  external  point  to  the  point  of  contact. 


EXERCISE  27 

1.  A  radius  that  bisects  an  arc  bisects  its  sub- 
tending chord  and  is  perpendicular  to  it. 

2.  On  a  circle  the  point  P  is  equidistant  from 
two  radii  OA  and  OB.  Prove  that  P  bisects  the 
arc  AB. 

3.  In  this  circle  the  chords  AM  and  ]\fB  are 
equal.  Prove  that  M  bisects  the  arc  AB  and  that 
the  radius  OM  bisects  the  chord  AB. 

4.  On  a  circle  are  five  points,  A,  B,  C,  D,  E,  so 
placed  that  .45,  BC,  CD,  DE  are  equal  chords. 
Prove  that  A  C,  BD,  CE  are  equal  chords,  and 
that  AD  and  BE  are  also  equal  chords. 

5.  If  two  chords  intersect  and  make  equal  angles 
with  the  diameter  through  their  point  of  intersec- 
tion, these  chords  are  equal. 


SECANTS  AND  TANGENTS  103 

Proposition  IX.    Theorem 

184.  A  line  perpendicular  to  a  radius  at  its  extrem- 
ity is  tangent  to  the  circle. 


A  p 

Given  a  circle,  with  XY  perpendicular  to  the  radius  OP  at  P. 
To  prove  that  XY  is  tangent  to  the  circle. 
Proof.    From  0  draw  any  other  line  to  XY,  as  OA. 
Then  OA>OP.  §86 

.'.  the  point  A  is  outside  the  circle.  §  160 

Hence  every  point,  except  P,  of  the  line  AT  is  outside  the 
circle. 

Therefore  AT  is  tangent  to  the  circle  at  P,  by  §  183.    q.e.d. 

185.  Corollary  1.   A  tangent  to  a  circle  is  perpendicular 

to  the  radius  drawn  to  the  point  of  contact. 

For  OP  is  the  shortest  line  from  O  to  XY,  and  is  therefore  ±  to  XY 
(§  86) ;  that  is,  XY  is  ±  to  OP. 

186.  Corollary  2.    A  perpendicular  to  a  tangent  at  the 

point  of  contact  passes  through  the  center  of  the  circle. 

For  a  radius  is  ±  to  a  tangent  at  the  point  of  contact,  and  therefore  a 
±  erected  at  the  point  of  contact  coincides  with  this  radius  and  passes 
through  the  center  of  the  circle. 

187.  Corollary  3.  A  perpendicular  from  the  center  of  a 
circle  to  a  tangent  passes  through  the  point  of  contact. 

What  does  §  86  say  about  this  perpendicular  ?  '•■     .- 


104 


BOOK  II.    PLANE  GEOMETEY 


188.  Concentric  Circles.    Two  circles  that  have  the  same  center 
are  said  to  be  concentric. 


EXERCISE  28 

1.  The  shortest  chord  that  can  be  drawn  through  a  given 
point  within  a  circle  is  that  which  is  perpendicular 
to  the  diameter  through  the  poiiit. 

Show  that  any  other  chord,  CD,  through  P,  is  nearer 
0  than  is  AB. 

2.  The  diameter  CD  bisects  the  arc  AB.  Prove 
that  ZCi^^  =ABAC. 

What  kind  of  a  triangle  is  A  ^  JBC  ? 

3.  Tangents  at  the  extremities  of  a  diameter 
are  parallel. 

4.  The  arc  AB  is  greater  than  the  arc  BC.  OP 
and  OQ  are  perpendiculars  from  the  center  to  AB 
and  BC  respectively.  Prove  that  Z  QPO  is  greater 
than  Z  OQP. 

5.  What  is  the  locus  of  the  center  of  a  circle  tangent  to  the 
line  XY  dX  the  point  P  ?    Prove  it. 

What  two  conditions  must  be  shown  to  be  fulfilled  ? 

6.  What  is  the  locus  of  the  mid-points  of  a  number  of  par- 
allel chords  of  a  circle  ?    Prove  it. 

7.  Three  equal  chords,  AB,  BC,  CD,  are  placed      /^^T^^^/? 
end  to  end,  and  the  radii  OA,  OB,  OC,  OD  are  2)/ 
drawn.    Prove  that  AAOC  =  AB0D. 

8.  All  equal  chords  of  a  circle  are  tangent  to  a 
concentric  circle. 

9.  If  a  number  of  equal  chords  are  drawn  in 
this  circle,  the  figure  gives  the  impression  of  a 
second  circle  inside  the  first  and  concentric  with 
it.    Explain  the  reason. 


SECANTS  AND  TANGENTS  105 

Proposition^  X.    Theorem 
189.  Two  parallel  lines  intercept  equal  arcs  on  a  circle. 


E-~--J!£^- -F  A- 


E 


Fig.  1  Fig.  2  Fig.  3 

Case  1.  When  the  parallels  are  a  tangent  and  a  secant  (Fig.  1). 
Given  AB^  a  tangent  at  P,  parallel  to  CD^  a  secant. 
To  prove  that  arc  CP  =  arc  DP. 

Proof.  Suppose  PP'  drawn  ±  to  AB  Sit  P. 

Then  PP'  is  a  diameter  of  the  circle.  §  186 

And  PP'  is  also  _L  to  CD.  §  97 

.-.  arc  CP  =  arc  Z)P.  §  174 

Case  2.  When  the  parallels  are  both  secants  (Fig.  2). 
Given  AB  and  CD,  parallel  secants. 
To  prove  that  arc  AC  =  arc  BD. 

Proof.    Suppose  EF  II  to  CD  and  tangent  to  the  circle  at  M. 
Then      arc  ^3/=  arc  BM,  and  arc  CM  =  arc  DM.        Case  1 
.-.  arc  ylC  =  arc  57).  Ax.  2 

Case  3.  When  the  parallels  are  both  tangents  (Fig.  3). 
Given  AB^  a  tangent  at  E,  parallel  to  CD^  a  tangent  at  F. 
To  prove  that        arc  FGE  =  arc  FHE. 
Proof.  Suppose  a  secant  GH  drawn  II  to  AB. 

Then       arc  GE  =  arc  HE,  and  arc  FG  =  arc  FH.        Case  1 
.'.  arc  FGE  =  arc  FHE,  by  Ax.  1.  Q. e.  d. 


106  BOOK  II.    PLANE  GEOMETRY 

Proposition  XI.    Theorem 

190.   TJirough  three  points  not  in  a  straight  line  one 
cii'cle,  and  only  one,  can  he  draion. 


Given  -A,  J?,  C,  three  points  not  in  a  straight  line. 

To  prove  that  one  circle,  and  only  one,  can  he  drawn  through 
A,  B,  and  C. 

Proof.  Draw  AB  and  BC. 

At  the  mid-points  of  AB  and  BC  suppose  Js  erected. 

These  -k  will  intersect  at  some  point  0,  since  AB  and  BC  are 
neither  parallel  nor  in  the  same  straight  line. 

The  point  O  is  in  the  perpendicular  bisector  of  AB,  and  is 
therefore  equidistant  from  A  and  B]  the  point  0  is  also  in 
the  perpendicular  bisector  of  BC,  and  is  therefore  equidistant 
from  B  and  C.  §  150 

Therefore  0  is  equidistant  from  A,  B,  and  C. 

Therefore  a  circle  described  about  0  as  a  center,  with  a 
radius  OA,  will  pass  through  the  three  given  points.       §  160 

The  center  of  any  circle  that  passes  through  the  three  points 
must  be  in  both  of  these  perpendicular  bisectors,  and  hence  at 
their  intersection.  As  two  straight  lines  can  intersect  in  only- 
one  point  (§  55),  0  is  the  only  point  that  can  be  the  center  of 
a  circle  through  the  three  given  points.  Q.  e.  d. 

191 .  Corollary.    Tivo  circles  can  intersect  in  only  two  points. 

If  two  circles  have  three  points  in  common,  can  it  be  shown  that  they 
coincide  and  form  one  circle  ? 


SECANTS  AND  TANGENTS  107 

Proposition  XII.    Theorem 

192.  Tlie  tangents  to  a  circle  draivn  from  an  external 
point  are  equal,  and  tnake  equal  angles  ivith  the  line 
joining  the  point  to  the  center. 


A 

Given  PA  and  P5,  tangents  from  P  to  the  circle  whose  center  is 
0,  and  PO  the  line  joining  P  to  the  center  0. 

To  prove  that     PA  =  PB,  and  ZAPO  =  Z  OPB, 
Proof.  Draw  OA  and  OB. 

PA  is  ±  to  OA,  and  PB  is  ±  to  OB.  §  185 

{A  tangent  to  a  circle  is  J_  to  the  radius  drawn  to  the  point  of  contact.) 

In  the  rt.  A  PAO  and  PBO, 

PO  =  PO,  Men. 

and                                          OA  =  OB.  §  162 

.-.  rt.  A  PA  0  is  congruent  to  rt.  A  PBO.  §  89 

.-.  PA  =  PB,  and  ZAPO  =  Z OPB,  by  §  67.  Q. e. d. 

193.  Line  of  Centers.  The  line  determined  by  the  centers  of 
two  circles  is  called  the  line  of  centers. 

194.  Tangent  Circles.  Two  circles  that  are  both  tangent  to  the 
same  line  at  the  same  point  are  called  tangent  circles. 

Circles  are  said  to  be  tangent  internally  or  externally,  according  as  they 
lie  on  the  same  side  of  the  tangent  Hne  or  on  opposite  sides.  E.g.  the 
two  circles  shown  in  the  figure  on  page  110  are  tangent  externally. 

The  point  of  contact  with  the  line  is  called  the  point  of  contact  or  point 
of  tamjency  of  the  circles. 


108 


BOOK  II.    PLANE  GEOMETRY 


EXERCISE  29 


1.  Show  that  the  reasoning  of  §  190  will  not  hold  for  four 
points,  and  hence  that  a  circle  cannot  always 
be  drawn  through  four  points. 


2.  Tangents  to  a  circle  3it  A,  B,  C,  points 
on  the  circle,  meet  in  P  and  Q,  as  here  shown. 
Prove  that  AP -\- QC  =  PQ. 

3.  If  a  quadrilateral  has  each  side  tangent  to 
a  circle,  the  sum  of  one  pair  of  opposite  sides 
equals  the  sum  of  the  other  pair. 

In  this  figure,  SP  -{■  QR  =  PQ  +  RS. 

4.  The  hexagon  here  shown  has  each  side 
tangent  to  the  circle.  Prove  that  AB-\-  CD + EF 
=  BC  -^  DE  -^  FA. 

5.  In  this  figure  CF  is  a  diameter  perpen- 
dicular to  the  parallel  chords  DB  and  EA,  and 
arc  AB  =  40°  and  arc  5C  =  50°.  How  many  de- 
grees are  there  in  arcs  CD,  DE,  EF,  and  FA  ? 

6.  In  this  figure  XYis  tangent  to  the  circle 
at  B,  the  chord  CA  is  perpendicular  to  the    j 
diameter  BD,  and  the  arc  CD  =  150°.    How 
many  degrees  are  there  in  arc  AB? 

7.  If  a  quadrilateral  has  each  side  tangent  to 
a  circle,  the  sum  of  the  angles  at  the  center 

subtended  by  any  two  opposite  sides  is  equal  to  a  straight  angle. 

S.  AP  and  CQ  are  parallel  tangents  meeting  a 
third  tangent  QP,  as  shown  in  the  figure.  0  be- 
ing the  center,  prove  that  the  angle  POQ  is  a 
right  angle. 

Are  A,  0,  and  C  in  the  same  straight  line  ?  Draw  OA 
and  OC,  and  find  the  relations  of  the  zi  at  0  to  those  at  P  and  Q. 


LINE  OF  CENTERS  109 

Proposition  XIII.    Theorem 

195.  If  tivo  circles  intersect,  the  line  of  centers  is  the 
perpendicular  bisector  of  their  common  chord. 


Given  0  and  0',  the  centers  of  two  intersecting  circles,  AB  the 
common  chord,  and  00'  the  line  of  centers. 

To  prove  that  00'  is  A.  to  AB  at  its  mid-point 
Proof.  Draw  OA,  OB,  O'A,  and  O'B. 

OA  =  OB,  and  O'A  =  O'B.  §  162 

.'.  0  and  O'  are  two  points,  each  equidistant  from  A  and  B. 
.*.  00'  is  the  perpendicular  bisector  of  AB,  by  §  151.     q.e.d. 

196.  Common  Tangents.  A  tangent  to  two  circles  is  called  a 
common  external  tangent  if  it  does  not  cut  the  line-segment 
joining  the  centers,  and  a  common  internal  tangent  if  it  cuts  it. 

EXERCISE  30 

Describe  the  relative  position  of  two  circles  if  the  line-segment 
joining  the  centers  is  related  to  the  radii  as  stated  in  Exs.  1-5, 
and  illustrate  each  case  hy  a  figure  : 

1.  The  line-segment  greater  than  the  sum  of  the  radii. 

2.  The  line-segment  equal  to  the  sum  of  the  radii. 

3.  The  line-segment  less  than  the  sum  but  greater  than  the 
difference  of  the  radii. 

4.  The  line-segment  equal  to  the  difference  of  the  radii. 

5.  The  line-segment  less  than  the  difference  of  the  radii. 


110  BOOK  11.    PLANE  GEOMETRY 

Proposition  XIV.    Theorem 

197.  If  two  circles  are  tangent  to  each  other,  the  line 
of  centers  passes  through  the  point  of  contact. 

A 


B 

Given  two  circles  tangent  at  P. 

To  prove  that  P  is  in  the  line  of  centers. 

Proof.    Let  AB  be  the  common  tangent  at  P.  §  194 

Then  a  J_  to  ^4  B,  drawn  through  the  point  P,  passes  through 

the  centers  0  and  0\  §  186 

{A  l.to  a  tangent  at  the  point  of  contact  passes  through  the  center 
of  the  circle.) 

Therefore  the  line  determined  by  0  and  0',  having  two  points 

in  common  with  this  _L,  must  coincide  with  it.  Post.  1 

.*.  P  is  in  the  line  of  centers.  q.e.d. 

EXERCISE  31 

Describe  the  relative  position  of  two  circles  having  tangents 
as  stated  in  Exs.  1~5,  and  illustrate  each  case  by  a  figure : 

1.  Two  common  external  and  two  common  internal  tangents. 

2.  Two  common  external  tangents  and  one  common  internal 
tangent. 

3.  Two  common  external  tangents  and  no  common  internal 
tangent. 

4.  One  common  external  and  no  common  internal  tangent. 

5.  No  common  tangent. 


TANGENTS  111 

6.  The  line  which  passes  through  the  mid-points  of  two 
parallel  chords  passes  through  the  center  of  the  circle. 

7.  If  two  circles  are  tangent  externally,  the  tangents  to  them 
from  any  point  of  the  common  internal  tangent  are  equal. 

8.  If  two  circles  tangent  externally  are  tangent  to  a  line 
AB  at  A  and  B,  their  common  internal  tangent  bisects  AB. 

9.  The  line  drawn  from  the  center  of  a  circle  to  the  point 
of  intersection  of  two  tangents  is  the  perpendicular  bisector  of 
the  chord  joining  the  points  of  contact. 

10.  The  diameters  of  two  circles  are  respectively  2.74  in.  and 
3.48  in.  Find  the  distance  between  the  centers  of  the  circles 
if  they  are  tangent  externally.  Find  the  distance  between  the 
centers  of  the  circles  if  they  are  tangent  internally. 

11.  Three  circles  of  diameters  4.8  in.,  3.6  in.,  and  4.2  in.  are 
externally  tangent,  each  to  the  other  two.  Find  the  perimeter 
of  the  triangle  formed  by  joining  the  centers. 

12.  A  circle  of  center  0  and  radius  r'  rolls  around  a  fixed 
circle  of  radius  r.    What  is  the  locus  of  O?    Prove  it. 

13.  The  line  drawn  from  the  mid-point  of  a  chord  to  the 
mid-point  of  its  subtended  arc  is  perpendicular  to  the  chord. 

14.  If  two  circles  tangent  externally  at  P  are  tangent  to  a 
line  AB  at  A  and  B,  the  angle  BPA  is  a  right  angle. 

15.  Three  circles  are  tangent  externally  at  the  points  A,  B, 
and  C,  and  the  chords  AB  and  AC  are  produced  to  cut  the 
circle  BC  Sit  D  and  E.    Prove  that  DE  is  a  diameter. 

16.  If  two  radii  of  a  circle,  at  right  angles  to  each  other, 
when  produced  are  cut  by  a  tangent  to  the  circle  at  A  and  B, 
the  other  tangents  from  A  and  B  are  parallel  to  each  other. 

17.  If  two  common  external  tangents  or  two  common  inter- 
nal tangents  are  drawn  to  two  circles,  the  line-segments  inter- 
cepted between  the  points  of  contact  are  equal. 


112  BOOK  11.    PLANE  GEOMETRY 

198.  Measure.  The  number  of  times  a  quantity  of  any  kind 
contains  a  known  unit  of  the  same  kind,  expressed  in  terms  of 
that  known  imit,  is  called  the  measure  of  the  quantity. 

Thus  we  measure  the  length  of  a  schoolroom  by  finding  the  number  of 
times  it  contains  a  known  unit  called  the  foot.  We  measure  the  area  of 
the  floor  by  finding  the  number  of  times  it  contains  a  known  unit  called 
the  square  foot.  You  measure  your  weight  by  finding  the  number  of 
times  it  contains  a  known  unit  called  the  pound.  Thus  the  measure  of 
the  length  of  a  room  may  be  30  ft.,  the  measure  of  the  area  of  the  floor 
may  be  600  sq.  ft.,  and  so  on. 

The  abstract  number  found  in  measuring  a  quantity  is  called 
its  numerical  tneasui^e,  or  usually  simply  its  measure. 

199.  Ratio.  The  quotient  of  the  numerical  measures  of  two 
quantities,  expressed  in  terms  of  a  common  unit,  is  called  the 
ratio  of  the  quantities. 

Thus,  if  a  room  is  20  ft.  by  35  ft.,  the  ratio  of  the  width  to  the  length 
is  20  ft,  -=-  35  ft.,  or  ||,  which  reduces  to  ^.    Here  the  common  unit  is  1  ft. 

The  ratio  of  a  to  6  is  written  -,  or  a  :  6,  as  in  arithmetic  and  algebra. 

b 
Thus  the  ratio  of  20°  to  30°  is  f§,  or  |,  or  2  :  3. 

200.  Commensurable  Magnitudes.  Two  quantities  of  the  same 
kind  that  can  both  be  expressed  in  integers  in  terms  of  a  com- 
mon unit  are  said  to  be  commensurable  magnitudes. 

Thus  20  ft.  and  35  ft.  are  expressed  in  integers  (20  and  35)  in  terms 
of  a  common  unit  (1  ft.);  similarly  2  ft.  and  3|  ft.,  the  integers  being 
4  and  7,  and  the  common  unit  being  \  ft. 

The  common  unit  used  in  measuring  two  or  more  commensurable 
magnitudes  is  called  their  common  measure.  Each  of  the  magnitudes  is 
called  a  multiple  of  this  common  measure. 

201.  Incommensurable  Magnitudes.  Two  quantities  of  the 
same  kind  that  cannot  both  be  expressed  in  integers  in  terms 
of  a  common  unit  are  said  to  be  incommensurable  magnitudes. 

Thus,  if  a  =  V2  and  6  =  3,  there  js  no  number  that  is  contained  an 
integral  number  of  times  in  both  V2  and  3.  Hence  a  and  h  are,  in  this 
case,  incommensurable  magnitudes. 


measureme:n^t  113 

202.  Incommensurable  Ratio.  The  ratio  of  two  incommensur- 
able magnitudes  is  called  an  incommensurable  ratio. 

Although  the  exact  value  of  such  a  ratio  cannot  be  expressed 
by  an  integer,  a  common  fraction,  or  a  decimal  fraction  of  a 
limited  number  of  places,  it  may  be  expressed  approximately. 

Thus  suppose  -  =  V2. 

Now  V2  =1.41421356  •••,  which  is  greater  than  1.414213 
but  less  than  1.414214.  Then  if  a  millionth  part  of  b  is  taken 
as  the  unit  of  measure,  the  value  oi  a:b  lies  between  1.414213 
and  1.414214,  and  therefore  differs  from  either  by  less  than 
0.000001. 

By  carrying  the  decimal  further  an  approximate  value  may 
be  found  that  will  differ  from  the  ratio  by  less  than  a  billionth, 
a  trilllonth,  or  any  other  assigned  value. 

That  is,  for  i^ractical  purposes  all  ratios  are  commensurable. 

For  example,  if  -  >  —  but  < ,  then  the  error  in  taking  either  of 

1)71  n 

these  values  for  -  is  less  than  - ,  the  difference  of  these  ratios.  But  by 

b  1  "" 

increasing  n  indefinitely,  -  can  be  decreased  indefinitely,  and  a  value  of 

the  ratio  can  be  found  within  any  required  degree  of  accuracy. 


EXERCISE  32 
Find  a  common  measure  of : 

1.  32  in.,  24  in.       3.  5^  in.,  3^  in.  5.  6^  da.,  2|  da. 

2.  48  ft,  18  ft.        4.  2|  lb.,  1^  lb.  6.  14.4  in.,  1.2  in. 

Find  the  greatest  common  measure  of: 

7.  64  yd.,  24  yd.     9.  7.5  in.,  1.25  in.      11.  2|  ft,  0.25  ft 

8.  51  ft.,  17  ft      10.  31  in.,  0.33J  in.     12.  75°,  7°  30'. 

13.  lia:b  =  V3,  find  an  approximate  value  of  this  ratio  that 
shall  differ  from  the  true  value  by  less  than  0.001. 


114 


BOOK  II.    PLANE  GEOMETKY 


203.  Constant  and  Variable.  A  quantity  regarded  as  having 
a  fixed  value  throughout  a  given  discussion  is  called  a  constant, 
but  a  quantity  regarded  as  having  different  successive  values 
is  called  a  variable. 

204.  Limit.  When  a  variable  approaches  a  constant  in  such 
a  way  that  the  difference  between  the  two  may  become  and 
remain  less  than  any  assigned  positive  quantity,  however 
small,  the  constant  is  called  the  Ihnit  of  the  variable. 

Variables  can  sometimes  reach  their  limits  and  sometimes  not.  E.g.  a 
chord  may  increase  in  length  up  to  a  certain  limit,  the  diameter,  and 
it  can  reach  this  limit  and  still  be  a  chord  ;  it  may  decrease,  approaching 
the  limit  0,  but  it  cannot  reach  this  limit  and  still  be  a  chord. 

205.  Inscribed  and  Circumscribed  Polygons.  If  the  sides  of  a 
polygon  are  all  chords  of  a  circle,  the  polygon  is  said  to  be 
inscribed  in  the  circle ;  if  the  sides  are  all  tangents  to  a  circle, 
the  polygon  is  said  to  be  circum- 
scribed about  the  circle. 

The  circle  is  said  to  be  circum- 
scribed about  the  inscribed  polygon, 
and  to  be  inscribed  in  the  circum- 
scribed polygon. 


Inscribed 
Polygon 


Circumscribed 
Polygon 


206.  Circle  as  a  Limit.  If  we  inscribe  a  square  in  a  circle, 
and  then  inscribe  an  octagon  by  taking  the  mid-points  of  the 
four  equal  arcs  for  the  new  vertices,  the  octa- 
gon is  greater  than  the  square  but  smaller  than 
the  area  inclosed  by  the  circle,  and  the  perim- 
eter of  the  octagon  is  greater  than  the  perim- 
eter of  the  square  (§  112). 

By  continually  doubling  the  number  of  sides 
in  this  way  it  appears  that  the  area  inclosed  by  the  circle  is  the 
limit  of  the  area  of  the  polygon,  and  the  circle  is  the  limit  of 
its  perimeter,  as  the  number  of  sides  is  indefinitely  increased. 

Hence  we  have  limiting  forms  as  well  as  limiting  values,  the  form  of 
the  circle  being  the  limit  approached  by  the  form  of  the  inscribed  polygon. 


LIMITS 


115 


-\L 


■\M 


207.  Principle  of  Limits.  If,  ivhile  approaching  their  respec- 
tive limits,  two  variables  are  always  equal,  their  limits  are  equal. 

Let  AX  and  BY  increase  in 
length  in  such  a  way  that  they 
always  remain  equal,  and  let 
their  respective  limits  be  ylL 
and  BM. 

To  prove  that  AL  =  BM. 

Suppose  these  limits  are  not  equal,  but  that  AZ  =  BM. 

Then  since  X  may  reach  a  point  between  Z  and  L  we  may 
have  AX>  AZ,  and  therefore  greater  than  its  supposed  equal, 
BM;  but  5 F cannot  be  greater  than  BM.  Therefore  we  should 
have  AX>  BY,  which  is  contrary  to  what  is  given. 

Hence  BM  cannot  be  greater  than  AL,  and  similarly  AL 
cannot  be  greater  than  BM.    .'.  AL  =  BM.  q.e.d. 

208.  Area  of  Circle.  The  area  inclosed  by  a  circle  is  called 
the  area  of  the  circle. 

It  is  evident  that  a  diameter  bisects  the  area  of  a  circle. 

209.  Segment.    A  portion  of  a  plane  bounded  by  an  arc  of  a 
circle  and  its  chord  is  called  a  segment 
of  the  circle. 

If  the  chord  is  a  diameter,  the  segment  is 
called  a  semicircle,  this  word  being  commonly 
used  to  mean  not  only  half  of  the  circle  but  also 
the  area  inclosed  by  a  semicircle  and  a  diameter. 

210.  Sector.     A    portion    of    a    plane  -^m,c.b- 
bounded  by  two  radii  and  the  arc  of  the  circle  intercepted  by 
the  radii  is  called  a  sector. 

If  the  arc  is  a  quarter  of  the  circle,  the  sector  is  called  a  quadrant. 

211.  Inscribed  Angle.  An  angle  whose  vertex  is  on  a  circle, 
and  whose  sides  are  chords,  is  called  an  inscribed  angle. 

An  angle  is  said  to  be  inscribed  in  a  segment  if  its  vertex  is  on  the  arc 
of  the  segment  and  its  sides  pass  through  the  ends  of  the  arc. 


116  BOOK  II.    PLANE  GEOMETRY 

Proposition  XV.    Theorem 

212.  In  the  same  circle  or  in  equal  circles  tivo  central 
angles  have  the  same  ratio  as  their  intercepted  arcs. 


Fig.  1  Fig.  2  Fig.  3 

Given  two  equal  circles  with  centers  0  and  0\  AOB  and  A'O'B' 
being  central  angles,  and  AB  and  A^B'  the  intercepted  arcs. 

To  prove  that  ,   ,  ^ —  = 

Case  1.  When  the  ares  are  commensurahle  (Figs.  1  and  2). 

Proof.    Let  the  arc  m  be  a  common  measure  of  A^B'  and  AB. 
Apply  the  arc  m  as  a  measure  to  the  arcs  A'B^  and  ^5  as 
many  times  as  they  will  contain  it. 

Suppose  m  is  contained  a  times  in  A'B',  and  b  times  in  AB. 


Then 


arc  A  *B'      a 


.SiVcAB       b 

At  the  several  points  of  division  on  AB  and  A 'B'  draw  radii. 

These  radii  will  divide  Z.AOB  into  b  j)arts,  and  Z.A'0'B' 

into  a  parts,  equal  each  to  each.  §  167 

ZA'O'B'      a 
ZAOB  ~1j' 

ZA'O'B'      slygA'B'    ,       ,       „ 

.-.      ■   ^  ^^  = — J  by  Ax.  8.  Q.E.D. 

ZAOB        SiYcAB       ^ 

Case  2  may  be  omitted  at  the  discretion  of  the  teacher  if  the  incom- 
mensurable cases  are  not  to  be  taken  in  the  course. 


MEASURE  OF  ANGLES  117 

Case  2.  When  the  arcs  are  incommensurable  (Figs.  2  and  3). 

Proof.  Divide  AB  into  a  number  of  equal  parts,  and  apply 
one  of  these  parts  to  A  'B'  as  many  times  as  A  'B'  will  contain  it. 

Since  AB  and  A'B'  are  incommensurable,  a  certain  number 
of  these  parts  will  extend  from  A'  to  some  point,  as  P,  leaving 
a  remainder  PB'  less  than  one  of  these  parts.    Draw  O'P. 

By  construction  AB  and  A'P  are  commensurable. 
ZA'O'P      arc^'P 


*  '  ZAOB       ^vgAB 


Case  1 


By  increasing  the  number  of  equal  parts  into  which  AB  is 
divided  we  can  diminish  the  length  of  each,  and  therefore  can 
make  PB'  less  than  any  assigned  positive  value,  however  small. 

Hence  PB'  approaches  zero  as  a  limit  as  the  number  of  parts 
of  AB  is  indefinitely  increased,  and  at  the  same  time  the 
corresponding  angle  PO'B'  approaches  zero  as  a  limit.     §  204 

Therefore  the  arc  ^'P  approaches  the  arc  A'B'  as  a  limit,  and 
the  A  A' O'P  approaches  the  Z  A'O'B'  as  a  limit. 

.  , ,    arc^'P  ,       arcyl'5'  ..     .^ 

.*.  the  variable -—  approaches — —  as  a  Innit, 

arc^^  arc  .45 

•  VI    AA'O'P  ,       Z.  A'O'B'  ..     .. 

and  the  variable     ,  ,  ,,„  approaches     .   .  __   as  a  limit. 
Z.AOB     ^^  AAOB 

ZA'O'P  .      ,  ,  ^    arc^'P 

But  .  ,  ^,,  IS  always  equal  to -—  > 

Z.AOB  ./       ^  siTcAB 

as  A'P  varies  in  value  and  approaches  A'B'  as  a  limit.     Case  1 

Z  A'O'B'      SiYcA'B'    ,      .  OAT  .^T^ 

.'.-^ = --— jby§207.  Q.E.D. 

ZAOB        SiTcAB       -^ 

213.  Numerical  Measure.  We  therefore  see  that  the  numerical 
measure  of  a  central  angle  (in  degrees,  for  example)  equals  the 
numerical  measure  of  the  intercepted  arc.  This  is  commonly 
expressed  by  saying  that  a  central  angle  is  measured  by  the 
intercepted  arc. 


118  BOOK  11.    PLANE  GEOMETRY 

Proposition  XVI.    Theorem 

214.  A7i  inscribed  angle  is  measured  hy  half  the  in- 
tercepted arc. 

B  B  B 


Given  a  circle  with  center  O  and  the  inscribed  angle  B^  inter- 
cepting the  arc  AC. 

To  prove  that  Z.  B  is  measured  hy  half  the  arc  A  C. 
Case  1.  When  0  is  on  one  side,  as  AB  (Fig.  1). 
Proof.  Draw  OC. 

Then                              '.'OC  =  OB,  §162 

.\AB  =  AC.  §74 

But                        Z.B-\-AC  =  Z.  A  OC.  §  111 

.•.2AB  =  AA0C.  Ax.  9 

.'.Z.B  =  :^  Z.AOC.  Ax.  4 

But                  AAOC  is  measured  by  arc  A C.  §  213 

.*.  1  Z.  AOC  is  measured  by  ^  arc  AC.  Ax.  4 

.'.  Z.B  is  measured  by  ^  arc  AC.  Ax.  9 

Case  2.  When  0  lies  within  the  angle  B  (Fig.  2). 

Proof.  Draw  the  diameter  BD, 

Then  Z  ABD  is  measured  by  \  arc  AD, 

and  Z  DBC  is  measured  by  ^  arc  DC.  Case  1 

.-.  Z  ^57)  +  Z  Z)5C  is  measured  by  \  (arc  ^i)  +  arc  i)C), 

or  Z  ^^C  is  measured  by  \  arc  AC. 


MEASURE  OF  ANGLES 


119 


Case  3.  W?ien  0  lies  outside  the  angle  B  (Fig.  3). 

Proof.  Draw  the  diameter  BD. 

Then  Z  DBC  is  measured  by  \  arc  7)C, 

and  Z  DBA  is  measured  by  \  arc  DA.  Case  1 

.'.  /.DBC  —  Z. DBA  is  measured  by  \  (arc  DC  —  arc  /).!), 
or 


Z.ABC  is  measured  by  \  arc  .4C. 


Q.E.D. 


Fig.  4 


Fig.  6 


215.  Corollary  1.  An  angle  inscribed  in  a  semicircle  is  a 
right  angle. 

For  it  is  half  of  a  central  straight  angle,  as,in  Fig,  4. 

216.  Corollary  2.  An  angle  inscribed  in  a  segment  greater 
than  a  semicircle  is  an  acute  angle,  and  an  angle  inscribed  in 
a  segment  less  than  a  semicircle  is  an  obtuse  angle. 

See  A  A  and  B  in  Fig,  5. 

217.  Corollary  3.  Angles  inscribed  in  the  same  segment 
or  in  equal  segments  are  equal. 

Why  is  this  ?  (Fig.  6.) 

218.  Corollary  A:.   If  a   quadrilateral  is  inscribed  in  a 

circle,  the  opposite  angles  are  supplementary ;  and,  conversely, 

if  two  opposite  angles  of  a  quadrilateral  are  supplementary, 

the  quadrilateral  can  be  inscribed  in  a  circle. 

For  the  second  part,  can  a  circle  be  passed  through  A^  /f^f/^X 

jB,  C  (§  190)  ?   If  it  does  not  pass  through  D  also,  can  you  f  I  d>--~^A 

show  that  Z  D  would  be  greater  than  or  less  than  some  other  jj^^^^^"^ 
angle  (§111)  that  is  supplementary  to  Z  S  ? 


120  BOOK  II.    PLAXE  GEOMETRY 

EXERCISE  33 

1.  A  parallelogram  inscribed  in  a  circle  is  a  rectangle. 

2.  A  trapezoid  inscribed  in  a  circle  is  isosceles. 

3.  The  shorter  segment  of  the   diameter   through  a  given 
point  within  a  circle  is  the  shortest  line  that  can 
be  drawn  from  that  point  to  the  circle.  f    o  p^ 

Let  P  be  the  given  point.  Prove  PA  shorter  than  any 
other  line  PX  from  P  to  the  circle. 

4.  The  longer  segment  of  the  diameter  through  a  given  point 
within  a  circle  is  the  longest  line  that  can  be  drawn  from  that 
point  to  the  circle. 

5.  The  diameter  of  the  circle  inscribed  in 
a  right  triangle  is  equal  to  the  difference 
between  the  hypotenuse  and  the  sum  of 
the  other  two  sides. 

6.  A  line  from  a  given  point  outside  a  circle  passing  through 
the  center  contains  the  shortest  line-segment  that  can  be  drawn 
from  that  point  to  the  circle. 

Let  P  be  the  point,  0  the  center,  A  the  point 
where  PO  cuts  the  circle,  and  C  any  other  point  on    ^' 
the  circle.    How  does  PC\-CO  compare  with  P02 

.  7.  A  line  from  a  given  point  outside  a  circle  passing  through 
the  center  contains  the  longest  line-segment  (to  the  concave 
arc)  that  can  be  drawn  from  that  point  to  the  circle. 

8.  Through  one  of  the  points  of  intersection  of  two  circles 
a  diameter  of  each  circle  is  drawn.    Prove  that 
the  line  joining  the  ends  of  the  diameters  passes 
through  the  other  point  of  intersection. 

9.  If  two  circles  intersect  and  a  line  is  drawn 
through  each  point  of  intersection  terminated  by 
the  circles,  the  chords  joining  the  corresponding 
ends  of  these  lines  are  parallel. 


MEASURE  OF  ANGLES  121 

Proposition  XVIT.    Theorem 

219.  A71  angle  formed  hy  two  chords  intersecting 
ivithin  the  circle  is  measured  hy  half  the  sum  of  the 
intercepted  arcs. 


Given  the  angle  AOB  formed  by  the  chords  AC  and  BD. 

To  prove  that  Z  A  OB  is  measured  hy  ^  (ar<?  AB  -f-  arc  CD^. 

Proof.        '  Draw  A  D. 

Then  AAOB  =  AA+Z.D.  §111 

{An  exterJ,or  Z  0/  a  A  is  equal  to  the  sum  of  the  two  opposite  interior  A.) 

But  Z  A  is  measured  by  ^  arc  CD,  §  214 

{An  inscribed  Z  is  measured  by  half  the  intercepted  arc.) 

and  Z  Z)  is  measured  by  ^  arc  AB.  §  214 

.-.  Z.  AOB  is  measured  by  J  (arc  A  B  +  arc  CD),  by  Ax.  1.    Q.  e.  d. 

Discussion.  If  0  is  at  the  center  of  the  circle,  to  what  previous  prop- 
osition does  this  proposition  reduce  ? 

If  0  is  on  the  circle,  as  at  U,  to  what  previous  proposition  does  this 
proposition  reduce  ? 

Suppose  the  point  0  remains  as  in  the  figure,  and  the  chord  AC 
swings  ahout  0  as  a  pivot  until  it  coincides  with  the  chord  BD.  What 
can  then  be  said  of  the  measure  of  A  AOB  and  COD  ?  What  can  be  said 
as  to  the  measure  oi  ABOC  and  DO  A  ? 

It  is  also  possible  to  prove  the  proposition  by  drawing  a  chord  AE 
parallel  to  BD,  and  showing  that  ZAOB  =  ZA,  since  they  are  alternate- 
interior  angles  formed  by  a  transversal  cutting  two  parallels.  Now  Z  A 
is  measured  by  I  arc  CE.  But  arc  CE  =  arc  CD  +  arc  DE^  or  arc  CD  + 
arc  ^B,  since  arc  AB  =  arc  DE  (§  189).  Therefore  ZAOB,  which  equals 
ZA,  is  measured  by  i  (arc  AB  -\-  arc  CD). 


122  BOOK  II.    PLAKE  GEOMETRY 

Propositiox  XVIII.    Theorem 

220.  An  angle  formed  hy  a  tangent  and  a  chord  drawn 
from  the  point  of  contact  is  measured  hy  half  the  in- 
tercepted arc. 


Given  the  chord  PQ  and  the  tangent  XY  through  P. 

To  prove  that  A  QPX  is  measured  hy  half  the  are  QSP. 

Proof.  Suppose  the  chord  QPi,  drawn  from  the  point  Q  par- 
allel to  the  tangent  XY. 

Then  arc  PA'  =  arc  QSP.  *      §  189 

{Two parallel  lines  intercept  equal  arcs  on  a  circle.) 

Also  Z  QPX  =  Z  PQR.  §  100 

{If  two  parallel  lines  are  cut  by  a  transversal,  the  alternate-interior 
angles  are  equal.) 

But  Z  PQR  is  measured  by  J  arc  PR.  §  214 

{An  inscribed  Z  is  measured  by  half  the  intercepted  arc.) 

Substitute  Z  QPX  for  its  equal,  the  Z  PQR, 
and  substitute  arc  QSP  for  its  equal,  the  arc  PR. 
Then  Z  QPX  is  measured  by  J  arc  QSP,  by  Ax.  9.         q.e.d. 

Discussion.  By  half  of  what  arc  is  Z  FPQ,  the  supplement  of  Z  QPX, 
measured  ? 

If  PQ  should  be  drawn  so  as  to  be  perpendicular  to  XY,  by  what 
would  A  YPQ  and  QPX  be  measured  ? 

Suppose  PQ  swings  about  the  point  P  as  a  pivot  until  it  coincides 
with  XY,  by  what  will  Z  YPQ  be  measured  ?  By  what  will  Z  QPX  be 
measured,  and  what  will  it  equal  ? 


MEASURE  OF  ANGLES 


123 


Proposition  XIX.    Theorem 

221.  An  angle  formed  hy  two  secants,  a  secant  and 
a  tangent,  or  tivo  tangents,  drawn  to  a  circle  from  an 
external  point,  is  measured  hy  half  the  difference  of 
the  intercepted  arcs. 


Fig.  1 


Fig.  3 


Given  two  secants  PBA  and  PCD^  from  the  external  point  P. 
To  prove  that  Z  P  is  measured  hy  ^  (arc  DA  —  arc  BC), 
Proof.    Suppose  the  chord  BX  drawn  II  to  PCD  (Fig.  1). 


Then 
Furthermore 


§189 


Ax.  9 
§102 
§214 


arc  BC  =  arc  DX. 
arc  XA  =  arc  DA  —  arc  DX. 
.'.  arc  XA  —  arc  DA  —  arc  BC. 
Also  ZP  =  ZXBA. 

But  Z  XBA  is  measured  by  ^  arc  XA. 
Substitute  AP  for  its  equal,  the  Z.XBA, 
and  substitute  arc  DA  —  arc  BC  for  its  equal,  the  arc  XA. 
Then  Z  P  is  measured  by  J  (arc  DA  —  arc  BC),  hy  Ax.  9.  Q.  e.  d. 
If  the  secant  PBA  Y  swings  around  to  tangency,  it  becomes 
the  tangent  PB  and  Fig.  1  becomes  Fig.  2.    If  the  secant  PCD 
also  swings  around  to  tangency,  it  becomes  the  tangent  PC  and 
Fig.  2  becomes  Fig.  3.    The  proof  of  the  theorem  for  each  of 
these  cases  is  left  for  the  student. 


124 


BOOK  11.    PLANE  GEOMETRY 


EXERCISE  34 

1.  If  two  circles  touch  each  other  and  two  lines  are  drawn 
through  the  point  of  contact  terminated  by  the  circles,  the  chords 
joining  the  ends  of  these  lines  are  parallel. 

This  could  be  proved  if  it  could  be  shown  that 
Z  A  equals  what  angle  ?  To  what  two  angles  can 
these  angles  be  proved  equal  by  §  220  ?  Are  those 
angles  equal  ? 

2.  If  one  side  of  a  right  triangle  is  the  diameter  of  a  circle, 
the  tangent  at  the  point  where  the  circle  cuts  the  hypotenuse 
bisects  the  other  side. 

If  OE  is  II  to  J.C,  then  because  BO  =  OA,  what  is  the 
relation  of  BE  to  EC  ?  The  proposition  therefore  reduces 
to  proving  that  OE  is  parallel  to  what  line  ?  This  can  be 
proved  if  Z  BOE  can  be  shown  equal  to  what  angle  ? 

3.  If  from  the  extremities  of  a  diameter  AD 
two  chords,  A  C  and  DB,  are  drawn  intersecting 
at  P  so  as  to  make  Z  APB  =  45°,  then  Z  BOC     ^ 
is  a  right  angle. 

4.  The  radius  of  the  circle  inscribed  in  an  equilateral  tri- 
angle is  equal  to  one  third  the  altitude  of  the  ^ 
triangle. 

To  prove  this  we  must  show  that  AF  equals  what 
line?  It  looks  as  if  AF  might  equal  EF,  and  EF 
equal  OF.  Is  there  any  way  of  proving  A  OFE  equi- 
lateral ?   of  proving  A  AEF  isosceles  ? 

5.  If  two  lines  are  drawn  through  any  point  in  a  diagonal 
of  a  square  parallel  to  the  sides,  the  points  where  these  lines 
meet  the  sides  lie  on  the  circle  whose  center 
is  the  point  of  intersection  of  the  diagonals. 

OY  =  OZ  if  what  two  A  are  congruent?  Why  are 
these  A  congruent  ?  OY  =  OX  if  what  two  A  are  con- 
gruent ?    OX  =  0W  if  what  two  A  are  congruent  ?  a 


PRINCIPLE  OF  CONTINUITY 


125 


222.  Positive  and  Negative  Quantities.     In 

geometry,  as   in  algebra,  quantities   may  be 
distinguished  as  positive  and  as  negative. 

Thus  as  we  consider  temperature  above  zero  posi- 
tive and  temperature  below  zero  negative,  so  in  this 
figure,  if  OB  is  considered  positive,  then  OD  may  be 
considered  negative.  Similarly,  if  OA  is  considered 
positive,  then  OC  may  be  considered  negative. 

Likewise  with  respect  to  angles  and  arcs,  if  the 
rotating  line  OA  moves  in  the  direction  of  AB, 
counterclockwise,  the  angle  and  arc  generated  are 
considered  positive.  If  it  rotates  in  the  direction 
AB',  like  the  hands  of  a  clock,  the  angle  and  arc 
generated  are  considered  negative. 

223.  Principle  of  Continuity.  By  considering  the  distinction 
between  positive  and  negative  magnitudes,  a  theorem  may 
often  be  so  stated  as  to  include  several  particular  theorems. 
For  example.  The  angle  included  between  two  lines  that  cut  or 
are  tangent  to  a  circle  is  measured  by  half  the  sum  of  the 
intercepted  arcs. 

In  particular  :  1.  If  the  lines  intersect  at  the  center,  half  the  sum  of 
the  arcs  will  then  become  simply  one  of  the  arcs,  and  the  proposition 
reduces  to  that  of  §  213. 

2.  If  the  lines  are  two  general  chords,  we  have  the  case  of  §  219. 

3.  If  the  point  of  in- 
tersection P  moves  to  the 

circle,  we  have  the  case     (  \/  \         f  2\    \         i  /P\ 
of  §  214,  one  arc  becom- 
ing zero. 

4.  If  P  moves  outside  the  circle,  then  the  smaller  arc  passes  through 
zero  and  becomes  negative,  so  that  the  sum  of  the  arcs  becomes  their 
arithmetical  difference  (§  221). 

We  may  continue  the  discussion  so  as  to  include  all  the  cases  of 
the  propositions  proved  from  §  213  to  §  221. 

When  the  reasoning  employed  to  prove  a  theorem  is  con- 
tinued as  just  illustrated,  so  as  to  include  several  theorems, 
we  are  said  to  reason  by  the  Principle  of  Continuity. 


126  BOOK  II.    PLANE  GEOMETRY 

224.  Problems  of  Construction.  At  the  beginning  of  the  study 
of  geometry  some  directions  were  given  for  simple  construc- 
tions, so  that  figures  might  be  drawn  with  accuracy.  It  was 
not  proved  at  that  time  that  these  constructions  were  correct, 
because  no  theorems  had  been  studied  on  which  proofs  could 
be  based.  It  is  now  purposed  to  review  these  constructions,  to 
prove  that  they  are  correct,  and  to  apply  the  methods  employed 
to  the  solution  of  more  difficult  problems. 

225.  Nature  of  a  Solution.  A  solution  of  a  problem  has  one 
requirement  that  a  proof  of  a  theorem  does  not  have. 

In  a  theorem  we  have  three  general  steps :  (1)  Given,  (2)  To 
prove,  (3)  Proof.  In  a  problem  we  have  four  steps  :  (1)  Given, 
(2)  Required  (to  do  some  definite  thing),  (3)  Construction  (show- 
ing how  to  do  it),  (4)  Proof  (that  the  construction  is  correct). 

We  prove  a  theorem,  but  we  solve  a  problem,  and  then  prove 
that  our  solution  is  a  correct  one. 

In  the  figures  of  this  text  given  lines  are  shown  as  full,  black  lines ; 
construction  lines  and  lines  required  are  shown  as  dotted  lines. 

226.  Discussion  of  a  Problem.  Besides  the  four  necessary 
general  steps  in  treating  a  problem,  there  is  a  desirable  step 
to  be  taken  in  many  cases.  This  is  the  discussion  of  the  prob- 
lem, in  which  is  considered  whether  there  is  more  than  one 
solution,  and  other  similar  questions. 

For  example,  suppose  the  problem  is  this  :  Required  from  a  given  point 
to  draw  a  tangent  to  a  circle. 

After  the  problem  has  been  solved  we  may  discuss  it  thus  :  In  general, 
if  the  given  point  is  outside  the  circle,  two  tangents  may  be  drawn, 
and  these  tangents  are  equal  (§  192) ;  if  the  given  point  is  on  the  circle 
only  one  tangent  can  be  drawn,  since  only  one  perpendicular  can  be 
drawn  to  a  radius  at  its  extremity  (§  184) ;  if  the  given  point  is  within 
the  circle,  evidently  no  tangent  can  be  drawn. 

In  the  discussion  the  Principle  of  Continuity  often  enters,  the  figure 
being  studied  for  various  positions  of  some  given  point  or  line,  as  was 
done  in  the  discussions  on  pages  121  and  122. 


PKOBLEMS  OF  CONSTRUCTION      127 

Proposition  XX.    Problem 

227.   To  let  fall  a  perpendicular  iq^on  a  given  line 
from  a  given  external  point. 

P\ 


Yy 


Given  the  line  AB  and  the  external  point  P. 
Required  from  P  to  let  fall  a  J_  upon  AB. 

Construction.    With  P  as  a  center,  and  a  radius  sufficiently 

great,  describe  an  arc  cutting  ^1^  at  J^  and  Y.  Post.  4 

With  X  and  Y  as  centers,  and  a  convenient  radius,  describe 

two  arcs  intersecting  at  C.  Post.  4 

Draw  PC.  Post.  1 

Produce  PC  to  intersect  AB  at  M.  Post.  2 

Then  PM  is  the  line  required.  Q.  e.  f. 

Proof.  Since  P  and  C  are  by  construction  two  points  each 
equidistant  from  X  and  F,  they  determine  the  perpendicular 
to  Jl  F  at  its  mid-point.  §  151 

{Two  points  each  equidistant  from  the  extremities  of  a  line  determine 

the  ±  bisector  of  the  line.)  Q.  E.  D. 

Discussion.   The  following  are  interesting  considerations  : 

That  PC  produced  will  really  intersect  AB,  as  stated  in  the  construc- 
tion, is  shown  in  the  proof. 

A  convenient  radius  to  take  for  the  two  intersecting  arcs  is  XY. 

If  C  falls  on  P,  take  C  at  the  other  intersection  of  the  arcs  below  AB^ 
as  is  seen  in  the  figure  of  Ex.  2,  p,  9. 

To  obtain  a  radius  for  the  first  circle,  draw  any  line  from  P  that  will 
cut  AB,  and  use  that. 


128  BOOK  II.    PLANE  GEOMETRY 

Proposition  XXI.    Problem 

228.  At  a  given  point  in  a  given  line,  to  erect  a  per- 
pendicular to  the  line, 

•>.a-  — -.  \y 


I 
!    \ 


Oy 
/ 


-^' 


^      \X       P         Yl    "  X--^^. 

Fig.  1  Fig.  2 

Given  the  point  P  in  the  line  AB. 
Required  to  erect  a  A.  to  AB  at  P. 

Case  1.  When  the  point  P  is  not  at  the  end  of  AB  (Fig.  1). 

Construction.    Take  PX  =  PY.  Post.  4 

With  X  and  Y  as  centers,  and  a  convenient  radius,  describe 

arcs  intersecting  at  C.  Post.  4 

Draw  CP.  Post.  1 

Then  CP  is  ±to  AB.  q.e.f. 

Proof.    P  and  C,  two  points  each  equidistant  from  X  and  Y, 
determine  the  ±  bisector  of  ZF,  by  §  151.  q.e.d. 

Case  2.  When  the  point  P  is  at  the  end  of  AB  (Fig.  2). 
Construction.    Suppose  P  to  coincide  with  B. 
Take  any  point  0  outside  of  AB,  and  with  0  as  a  center  and 
OB  as  a  radius  describe  a  circle  intersecting  AB  a,t  X. 

From  X  draw  the  diameter  XY,  and  draw  BY.  Post.  1 

Then  i^F  is  ±  to  AB.  Q.e.f. 

Proof.  Z  5  is  a  right  angle.  §  215 

.-.  BY  is  ±  to  AB,  by  §  27.  Q.e.d. 

Discussion.   If  the  circle  described  with  0  as  a  center  is  tangent  to 
AB  ?it  B,  then  OB  is  the  required  perpendicular  (§185). 


PROBLEMS  OF  CONSTRUCTION 

Proposition  XXII.    Problem 
229.   To  bisect  a  given  line. 

I 


129 


■B 


M 


Given  the  line  AB. 
Required  to  bisect  AB. 

Construction.    With  A  and  B  as  centers  and  AB  as  a  radius 

describe  arcs  intersecting  at  X  and  Y,  and  draw  XY.    Post.  4 

Then  XY  bisects  AB.  q.e.f. 

Proof.  XY  bisects  AB,  by  §  151.  q.e.d. 


Proposition  XXIII.    Problem 
230.   To  bisect  a  given  arc. 

\g 


Given  the  arc  AB. 

Required  to  bisect  AB. 

Construction.    Draw  the  chord  AB.  Post.  1 

Draw  CM,  the  perpendicular  bisector  of  the  chord  AB.    §  229 


Proof. 


Then  CM  bisects  the  arc  AB. 
CM  bisects  the  arc  AB,  by  §  177. 


Q.E.F. 
Q.E.D. 


130  BOOK  11.    PLANE  GEOMETRY 

Proposition  XXIV.    Problem 
231.   To  bisect  a  given  angle. 


o 

Given  the  angle  AOB. 
Required  to  bisect  A  AOB. 

Construction.    With  O  as  a  center  and  any  radius  describe  an 

arc  cutting  OA  at  X  and  OB  at  Y.  Post.  4 

With  X  and  Y  as  centers  and  .Y  F  as  a  radius  describe  arcs 

intersecting  at  P.  Post.  4 

Draw  OP.  Post.  1 

Then  OP  bisects  Z.AOB.  q.e.f. 

Proof.  Draw  PX  and  PY. 

Then  prove  that  the  A  OXP  and  0  YP  are  congruent.       §  80 
Then  AA0P  =  Z.P0B,hj4  67.  Q. e. d. 

EXERCISE  35 

1.  To  construct  an  angle  of  45°;  of  135°. 

2.  To  construct  an  angle  of  22°  30';  of  157°  30'. 

3.  To  construct  an  equilateral  triangle,  having  given  one 
side,  and  thus  to  construct  an  angle  of  60°. 

4.  To  construct  an  angle  of  30° ;  and  thus  to  trisect  a  right 
angle. 

5.  To  construct  an  angle  of  15°;  of  7°  30';  of  195°;  of  345°. 

6.  To  construct  a  triangle  having  two  of  its  angles  equal 
to  75°.    Is  the  triangle  definitely  determined  ? 


PROBLEMS  OF  CONSTRUCTIOISr  131 

Proposition  XXV.    Problem 

232.  From,  a  given  point  in  a  given  line,  to  draw  a 
line  making  an  angle  equal  to  a  given  angle. 


p- 

Ic 

\m       * 

Given  the  angle  AOB  and  the  point  P  in  the  line  PQ. 

Required  from  P  to  draw  a  line  making  with  the  line  PQ 
an  angle  equal  to  A  A  OB. 

Construction.  With  O  as  a  center  and  any  radius  describe  an 
arc  cutting  OA  at  C  and  OB  at  D.  Post.  4 

With  P  as  a  center  and  the  same  radius  describe  an  arc  MX^ 
cutting  PQ  at  M,  Post.  4 

With  M  as  a  center  and  a  line  joining  C  and  D  as  a  radius 
describe  an  arc  cutting  the  arc  MX  at  N.  Post.  4 

Draw  PN.  Post.  1 

Then  Z  QPN  =  Z  A  OB.  Q.  e.  f. 

Proof.  Draw  CD  and  MN. 

Then  prove  that  the  A  PMN  and  OCD  are  congruent.     §  80 
Then  Z  QPN  =Z.A  OB,  by  §  67.  Q.  e.  d. 

233.  Corollary.  Through  a  given  external  point,  to  draiv 
a  line  parallel  to  a  given  line. 

X 

Let  AB  be  the  given  hne  and  P  the  given  external  „/' 

point.  C -f-^-D 

Draw  any  Une  XPY  through  P,  cutting  AB  as  in  /q 

the  figure.  / 

Draw  CD  through  P,  making  Z.p  =  Zq.  / 

The  Hne  CD  will  be  the  line  required. 


132  BOOK  II.    PLANE  GEOMETRY 

Proposition  XXVI.    Problem 

234.  To  divide  a  given  line  iiito  a  given  number  of 
equal  parts. 

Ai^ 1 ; iB 

/^     7  7  7 

'  ^^-.  / 

Given  the  line  AB. 

Required  to  divide  AB  into  a  given  number  of  equal  parts. 

Construction.  From  A  draw  the  line  AO,  making  any  con- 
venient angle  with  AB.  Post.  1 

Take  any  convenient  length,  and  by  describing  arcs  apply 
it  to  ^  0  as  many  times  as  is  indicated  by  the  number  of  parts 
into  which  ^^  is  to  be  divided.  Post.  4 

From  C,  the  last  point  thus  found  on  A  0,  draw  CB.     Post.  1 

From  the  division  points  on  ^  0  draw  parallels  to  CB.  §  233 
These  lines  divide  AB  into  equal  parts.  q.e.f. 

Proof.    These  lines  divide  AB  into  equal  parts,  by  §  134.  q.  e.  d. 

EXERCISE  36 

1.  To  divide  a  given  line  into  four  equal  parts. 

2.  To  construct  an  equilateral  triangle,  given  the  perimeter. 

3.  Through  a  given  point,  to  draw  a  line  which  shall  make 
equal  angles  with  the  two  sides  of  a  given  angle. 

4.  Through  a  given  point,  to  draw  two  lines  so  that  they 
shall  form  with  two  intersecting  lines  two  isosceles  triangles. 

5.  To  construct  a  triangle  having  its  three  angles  respec- 
tively equal  to  the  three  angles  of  a  given  triangle. 


PKOBLEMS  OF  CONSTRUCTION  133 

Proposition  XXVII.    Problem 

235.   To  construct  a  triangle  lohen  two  sides  and  the 
included  angle  are  given. 


4. :i— X 

;c  B 


Given  h  and  c  two  sides  of  a  triangle,  and  O  the  included  angle. 
Required  to  construct  the  triangle. 

Construction.    On  any  line  as  AX,  by  describing  an  arc,  mark 

o^  AB  equal  to  c.  Post.  4 

At  A  construct  A  BAD  equal  to  Z  0.  §  232 

On  AD,  by  describing  an  arc,  mark  off  ^  C  equal  to  h.  Post.  4 

Draw  BC.  Post.  1 

Then  A  ABC  is  the  A  required.  q.e.f. 

Proof.    (Left  for  the  student.) 

236.  Corollary  1.    To  construct  a  triangle  when  a  side  and 

two  angles  are  given. 

There  are  two  cases  to  be  considered:  (1)  when  the  given  side  is 
included  between  the  given  angles ;  and  (2)  when  it  is  not  (in  which 
case  find  the  other  angle  by  §  107). 

237.  Corollary  2.  To  construct  a  triangle  when  the  three 
sides  are  given. 

238.  Corollary  3.  To  construct  a  parallelogram  when  two 
sides  and  the  included  angle  are  given. 

Combine  §  235  and  §  233. 


134  BOOK  II.    PLANE  GEOMETRY 

Proposition  XXVIII.    Problem 

239.   To  construct  a  triangle  ivheji  two  sides  and  the 
angle  opposite  one  of  them  are  given. 


yY 


O/' 




Given  a  and  h  two  sides  of  a  triangle,  and  A  the  angle  opposite  a. 
Required  to  construct  the  triangle. 
Construction.    Case  1.   If  a  is  less  than  h. 

Construct  Z  XA  Y  equal  to  the  given  A  A.  §  232 

On  A  Y  take  A  C  equal  to  h. 
Prom  C  as  a  center,  with  a  radius  equal  to  a,  describe  an  arc 
intersecting  the  line  AX  dX  B  and  B\ 

Draw  EC  and  B'C,  thus  completing  the  triangle. 

Then  both  the  ^ABC  and  AB^C  satisfy  the  conditions,  and 

hence  we  have  two  constructions.  q.e.f. 

This  is  called  the  ambiguous  case. 

Discussion.    If  the  given  side  a  is  equal 
to  the  ±  CB,  the  arc  described  from  C  will  yi 

touch  AX,  and  there  will  be  but  one  con- 
struction, the  rt.  A  ABC. 

If  the  given  side  a  is  less  than  the  per- 
pendicular from  C,  the  arc  described  from  ^ 
C  will  not  intersect  or  touch  AX,  and  hence  yY 
a  construction  is  impossible.                                                      (j  / 

If  Z  A  is  right  or  obtuse,  a  construe-  y^  \  ^ 

tion  is  impossible,  since  a<h\   for  the  side  X        \   ^ 

of  a  triangle  opposite  a  right  or  obtuse  angle  X 

is  the  longest  side  (§114).  A 


.Y 


PROBLEMS  OF  CONSTRUCTION      135 

Case  2.   If  a  is  equal  to  b. 

If  the  given  Z.  A  is  acute,  and  a  =  b,  the  arc  described 
from  C  as  a  center,  and  with  a  radius  equal  to  a,  will  cut 
the    line    WX   at   the    points    A    and    B.  y 

There  is  therefore  but  one  triangle  that  ?<^ 

satisfies  the  conditions,  namely  the  isos-    ^^ :^ll'____  \/  _  v- 

celes  A  ABC.  ^^^- — -''^ 

Discussion.  If  the  ZA  is  right  or  obtuse,  a  construction  is  impossible 
when  a  =  b  ;  for  equal  sides  of  a  triangle  have  equal  angles  opposite  them, 
and  a  triangle  cannot  have  two  right  angles  or  two  obtuse  angles  (§  109). 

Case  3.    If  a  is  greater  than  h. 

If  the  given  Z  A   is  acute,  the  arc  described  from  C  will 

cut  the  line  WX  on  opposite  sides  of  .1,  at  B  and  B'.    The 

A  ABC   satisfies    the   conditions,   but    the   A  AB'C  does   not, 

for  it  does  not  contain  the  acute  Z.A. 

C  "^ 
There   is   then   only   one   triangle   that  •      a^y/^\a     ] 

satisfies    the    conditions,    namely    the  ^.-.l\grS/. ^:>J...x 

A  ABC.  ^'^^< '^^'^ 

If  the  given  Z  .1  is  right,  the  arc  described 
from  C  cuts  the  line  WX  on  opposite  sides  of  A 

A  at  the  points  B  and  B',  and  we  have  the       x^    /    j^>\    / 
two  congruent  right  triangles  ABC  and  AB'C  ^^  "b^"^~~"^'b''^ 
that  satisfy  the  conditions. 

If  the  given  Z  .1  is  obtuse,  the  arc  de-  \^ 

scribed    from  C  cuts    the    line  WX  on  ,      n/  \\a     , 

oj)posite  sides  of  A,  at  the  points  B  and    -^y i^'l i..i^i.-x 

B'.    The  A  ABC  satisfies  the  conditions, 

but  the  A  AB'C  does  not,  for  it  does  not  contain  the  obtuse 
Z.A.  There  is  then  only  one  triangle  that  satisfies  the  con- 
ditions, namely  the  A  ABC, 

Discussion.  We  therefore  see  that  when  a  >  6,  we  have  only  one 
triangle  that  satisfies  the  conditions,  for  the  two  congruent  right  tri- 
angles give  us  only  one  distinct  triangle. 


136  BOOK  11.    PLANE  GEOMETRY 

Proposition  XXIX.    Problem 
240.   To  circumscribe  a  circle  about  a  given  triangle. 

^^ ^c 


Given  the  triangle  ABC. 

Required  to  circumscribe  a  O  about  A  ABC. 

Construction.  Draw  the  perpendicular  bisectors  of  the  sides 
^^and^C.  §229 

Since  AB  is  not  the  prolongation  of  CA,  these  Js  will  inter- 
sect at  some  point  0.  Otherwise  they  would  be  II,  and  one  of 
them  would  have  to  be  ±.  to  two  intersecting  lines.  §  82 

With  0  as  a  center,  and  a  radius  OA,  describe  a  circle.    Post.  4 

The  O^^C  is  the  O  required.  q.e.f. 

Proof.  The  point  0  is  equidistant  from  A  and  B,  and  also  is 
equidistant  from  A  and  C.  §  150 

.*.  the  point  0  is  equidistant  from  A,  B,  and  C. 
.*.  a  O  described  with  0  as  a  center,  with  a  radius  equal  to  OA, 
will  pass  through  the  vertices  A,  B,  and  C,  by  §  160.         q.e.d. 

241.  Corollary  1.  To  describe  a  circle  through  three  points 
not  in  the  same  straight  line. 

242.  Corollary  2.  To  find  the  center  of  a  given  circle  or 
of  the  circle  of  which  an  arc  is  given. 

243.  Circumcenter.  The  center  of  the  circle  circumscribed 
about  a  polygon  is  called  the  circumcenter  of  the  polygon. 


PROBLEMS  OF  CONSTEUCTION 

Proposition  XXX.    Problem 
244.   To  inscribe  a  circle  in  a  given  triangle. 


137 


P  B 

Given  the  triangle  ABC. 

Required  to  inscribe  a  O  in  A  ABC. 

Construction.  Bisect  the  A  A  and  B.  §  231 

From  0,  the  intersection  of  the  bisectors,  draw  OP  1.  to  the 
side  AB.  §  227 

With  0  as  a  center  and  a  radius  OP,  describe  the  O  PQR. 

The  OPQR  is  the  O  required.  q.e.f. 

Proof.  Since  0  is  ii>  the  bisector  of  the  Z.A,it  is  equidistant 
from  the  sides  AB  and  AC;  and  since  0  is  in  the  bisector  of 
the  Z.B,  it  is  equidistant  from  the  sides  AB  and  BC.        §  152 

.'.a  circle  described  with  0  as  a  center,  and  a  radius  OP, 
will  touch  the  sides  of  the  triangle,  by  §  184.  q.e.d. 

245.  Incenters  and  Excenters.  The  center  of  a  circle  inscribed 
in  a  polygon  is  called  the  incenter  of  the  polygon. 

The  intersections  of  the  bisectors 
of  the  exterior  angles  of  a  triangle 
are  the  centers  of  three  circles,  each 
tangent  to  one  side  of  the  triangle 
and  the  two  other  sides  produced. 
These  three  circles  are  called  escribed 
circles ;  and  their  centers  are  called 
the  excenters  of  the  triangle. 


138 


BOOK  II.    PLANE  GEOMETRY 


Proposition  XXXI.    Problem 

246.   Through  a  given. point,  to  draw  a  tangent  to  a 
given  circle. 


. — .P 


--^  M,--' 


-^''.P 


Fig.  1 

Given  the  point  P  and  the  circle  with  center  0. 
Required  through  P  to  draw  a  tangent  to  the  circle. 

Case  1.  When  the  given  point  is  on  the  circle  (Fig.  1). 

Construction.    From  the  center  0  draw  the  radius  OP.    Post.  1 

Through  P  draw  AT  _L  to  OP.  §  228 

Then  AF  is  the  tangent  required.  q.e.f. 

Proof.  Since  AF  is  ±  to  the  radius  OP,  Const. 

.*.  AF  is  tangent  to  the  O  at  P,  by  §  184.  q.e.d. 

Case  2.  When  the  given  point  is  outside  the  circle  (Fig.  2). 

Construction.    .  Draw  OP.  Post.  1 

Bisect  OP.  §  229 

With  the  mid-point  of  OP  as  a  center  and  a  radius  equal  to 
I"  OP,  describe  a  circle  intersecting  the  given  circle  at  the 
points  M  and  N,  and  draw  PM. 

Then  PM  is  the  tangent  required.  Q.  e.  f. 

Proof.  Draw  OM. 

Z  OMP  is  a  right  angle.  §  215 

.'.PilfisJ- to  OJ/.  !27 

.*.  PM  is  tangent  to  the  circle  at  M,  by  §  184.       q.e.d. 

Discussion.   In  like  manner,  we  may  prove  PN  tangent  to  the  given  O. 


PEOBLEMS  OF  CONSTRUCTION  139 

Proposition  XXXII.    Problem 

247.  Upon  a  given  line  as  a  chord,  to  describe  a  segment 
of  a  circle  in  which  a  given  angle  may  he  inscribed. 

,'        /o    \  \    / 

\     /     !    ""v  '.  ly 


Given  the  line  AB  and  the  angle  m. 

Required  on  AB  as  a  chords  to  describe  a  segment  of  a  circle 
in  which  Z.  m  may  be  inscribed. 

Construction.    Construct  the  ZABX  equal  to  the  Z.m.   §  232 

Bisect  the  line  AB  by  the  ±  PO.  §  229 

From  the  point  B  draw  BO  _L  to  XB.  §  228 

•  With  0,  the  point  of  intersection  of  PO  and  BO,  as  a  center, 
and  a  radius  equal  to  OB,  describe  a  circle. 

The  segment  ABQ  is  the  segment  required.         q.e.f. 

Proof.    The  point  0  is  equidistant  from  A  and  B.  §  150 

.'.  the  circle  will  pass  through  A  and  B.  §  160 

But  BX  is  ±  to  OB.  Const. 

.'.  BX  is  tangent  to  the  O.  §  184 

.'.  AABX  is  measured  by  ^  arc  ^5.  §  220 

But  any  angle,  as  the  Z  Q,  inscribed  in  the  segment  ABQ  is 

measured  by  ^  a,vcAB.  §  214 

.\ZQ  =  ZABX.  Ax.  8 

But  Z.ABX  =  Zm.  Const. 

.'.  Z.m  may  be  inscribed  in  the  segment  ABQ,  by  §  217.     Q.  e.d. 


140  BOOK  11.    PLANE  GEOMETRY 

248.  How  to  attack  a  Problem.  There  are  three  common 
methods  by  which  to  attack  a  new  problem : 

(1)  By  synthesis ; 

(2)  By  analysis ; 

(3)  By  the  intersection  of  loci. 

249.  Synthetic  Method.  If  a  problem  is  so  simple  that  the 
solution  is  obvious  from  a  known  proposition,  we  have  only  to 
make  the  construction  according  to  the  proposition,  and  then 
to  give  the  synthetic  proof,  if  a  proof  is  necessary,  that  the 
construction  is  correct. 

It  is  rarely  the  case,  however,  that  a  problem  is  so  simple  as  to  allow 
this  method  to  be  used.  We  therefore  commonly  resort  at  once  to  the 
second  method. 

250.  Analytic  Method.  This  is  the  usual  method  of  attack, 
and  is  as  follows  : 

(1)  Suppose  the  problem  solved  and  see  what  results  follow. 

(2)  Then  see  if  it  is  possible  to  attain  these  results  and  thus 
effect  the  required  construction;  in  other  words,  try  to  work 
backwards. 

The  third  method,  by  the  intersection  of  loci,  is  considered  on  page  143. 

251.  Determinate,    Indeterminate,    and   Impossible    Cases.    A 

problem  that  has  a  definite  number  of  solutions  is  said  to  be 
determinate.  A  problem  that  has  an  indefinite  number  of  solu- 
tions is  said  to  be  indeterminate.  A  problem  that  has  no  solu- 
tion is  said  to  be  impossible. 

For  example,  to  construct  a  triangle,  having  given  its  sides,  is  deter- 
minate ;  to  construct  a  quadrilateral,  having  given  its  sides,  is  indetermi- 
nate ;  to  construct  a  triangle  with  sides  2  in.,  3  in.,  and  6  in.  is  impossible. 

252.  Discussion.  The  examination  of  a  problem  with  refer- 
ence to  all  possible  conditions,  particularly  with  respect  to  the 
number  of  solutions,  is  called  the  discussion  of  the  problem. 

Discussions  have  been  given  in  several  of  the  preceding  problems. 


SOLUTION  OF  PKOBLEMS 


141 


253.  Applications  of  the  Anal3rtic  Method.    The  following  are 
examples  of  the  use  of  analysis  in  the  solution  of  problems. 


EXERCISE  37 

1.  In  a  triangle  ABC,  to  draw  PQ  parallel  to  the  base  AB, 
cutting  the  sides  in  P  and  Q,  so  that  PQ  shall  equal  AP  -\-  BQ. 

Analysis.   Assume  the  problem  solved. 

Then  AP  must  equal  some  part  of  PQ,  as  PX, 
and  BQ  must  equal  QX. 

But  if  AP  =  PX,  what  must  Z  PXA  equal  ? 

•.•  PQ  is  II  to  AB,  what  does  Z  PXA  equal  ? 

Then  why  must  ZBAX  =  ZXAP? 

Similarly,  what  about  ZQBX  and  Z XBA  ? 

Construction.  Now  reverse  the  process.  "What  should  we  do  to  zi  yl  and 
B  in  order  to  fix  X  ?   Then  how  shall  PQ  be  drawn  ?   Now  give  the  proof. 

2.  To  construct  a  triangle,  having  given  the  perimeter,  one 
angle,  and  the  altitude  from  the  vertex  of  the  given  angle. 

Analysis.  Let  ABC  hQ  the  triangle,  Z  G  the  given  angle,  and  CP 
the  given  altitude,  and  assume  that  the  problem  is  solved. 

Since   the   perimeter  is 
given  as  a  definite  line,  we    X- 
now  try  producing  AB  and 
BA,  making  BN=BC,  and 
AM=AC. 

Then  Zm  =  what  angle, 
and  Zn  =  what  angle  ? 

Then  Zm-\-  Zn-\-  Z MCN  =  180°. 

But  ZMCN=Zm'-^ZACB-\-Zn\ 

.'.  2 Zm  +  2 Zn-^ZACB  =  180°.    (Why?) 
.-.  Zm  +  Zn  +  IZACB  =  90°, 
or  Zm-{-  Zn  =  90°-iZACB. 

.:  Z  MCN  =90°  -\-^ZA  CB.    (Why  ?) 
.-.  Z  MCN  is  known. 

Construction.  Now  reverse  the  process.  Draw  MN  equal  to  the  perim- 
eter. Then  on  MN  construct  a  segment  in  which  Z  MCN  may  be  inscribed 
(§  247).  Draw  XC  II  to  MN  at  the  distance  CP  from  MN,  cutting  the  arc 
at  C.  Then  A  and  B  are  on  the  ±  bisectors  of  CM  and  CN.  Why  ? 


PB 


142  BOOK  II.    PLANE  GEOMETRY 

3.  To  draw  through  two  sides  of  a  triangle  a  line  parallel 
to  the  third  side,  so  that  the  part  intercepted  c  ^ 
between  the  sides  shall  have  a  given  length. 

If  PQ  =  d,  what  does  AR  equal  ?    How  will  you 
reverse  the  reasoning  ?  A        R  ~b 

4.  To  draw  a  tangent  to  a  given  circle  so  that  it  shall  be 
parallel  to  a  given  line. 

5.  To  construct  a  triangle,  having  given  a  side,  an  adjacent 
angle,  and  the  difference  of  the  other  sides. 

If  AB^  ZA,  and  AC  —  BC  are  known,  what  points  are 
determined  ?  Then  can  XB  be  drawn  ?  What  kind  of  a  tri-    -^ 
angle  is  A  XBC  ?   How  can  C  be  located  ?  ^  « 

6.  To  construct  a  triangle,  having  given  two  angles  and 
the  sum  of  two  sides.  ^ 

Can  the  third  Z  be  found  ?  Assume  the  prob- 
lem solved.  If  AX  =  AB  +  BC,  what  kind  of  a 
triangle  is  A  BXC  ?  What  does  Z  CBA  equal  ? 
Is  Z  X  known  ?   How  can  C  be  fixed  ? 

7.  To  construct  a  square,  having  given  the  diagonal. 

8.  To  draw  through  a  given  point  P  between  the  sides  of 
an  angle  AOB  ?i  line  terminated  by  the  sides  ^ 

of  the  angle  and  bisected  at  P.  / 

If  PM=:  PN,  and  PR  is  II  to  AO,  what  can  you    2?/>_>p 

say  as  to  OR  and  RN?   Can  you  now  reverse  this?    /  

Similarly,  if  PQ  is  II  to  BO,  is  OQ  =  to  QM?  ^        ^        MA 

9.  To  draw  a  line  that  would  bisect  the  angle  formed  by 
two  lines  if  those  lines  were  produced  to  meet. 

li  AB  and  CD  are  the  given  lines,  consider  what  would  be  the  con- 
ditions if  they  could  be  produced  to  meet  at  0.   Then  the  q 
bisector  of  Z  O  would  be  the  ±  bisector  of  PQ,  a  line  drawn              /\s^ 
so  as  to  make  equal  angles  with  the  two  given  lines.                   /  I  \ 

Now  reverse  this.    How  can  we  draw  PQ  so  as  to  make         /    j     >^ 
ZP  =  ZQ?   Draw  BR  II  to  DC,  and  lay  off  BR  =  BQ.    ^ 
Then  draw  QRP  and  prove  that  this  is  such  a  line.  Then 
draw  its  _L  bisector. 


EXERCISES  IN  LOCI  14B 

254.  Intersection  of  Loci.  The  third  general  method  of  attack 
mentioned  in  §  248  is  by  intersection  of  loci.  This  is  very  con- 
venient when  we  wish  to  find  a  point  satisfying  two  conditions, 
each  of  which  involves  some  locus. 


EXERCISE  38 

1.  To  find  a  point  that  is  i  in.  from  a  given  point  and  ^\  in. 
from  a  given  line.  ---^'-— _w_ 

If  P  is  the  given  point,  what  is  the  locus  of  a — I- ; ^ B 

a  point  I  in.  from  P  ?    If  ^^  is  the  given  line,  >      ^      l__ 

what  is  the  locus  of  a  point  y\  in.  from  AB  ?  --—■-' 

These  two  loci  intersect  in  how  many  points  at  most?    Discuss  the 

solution. 

2.  To  find  a  point  that  is  J  in.  from  one  given  point  and  |  in. 
from  another  given  point. 

Discuss  the  number  of  possible  points  answering  the  conditions. 

3.  To  find  a  point  that  is  \  in.  from  the  vertex  of  an  angle 
and  equidistant  from  the  sides  of  the  angle. 

4.  To  find  a  point  that  is  equidistant  from  two  intersecting 
lines  and  \  in.  from  their  point  of  intersection. 

How  many  such  points  can  always  be  found  ? 

5.  To  find  a  point  that  is  \  in.  from  a  given  point  and  equi- 
distant from  two  intersecting  lines. 

Discuss  the  problem  for  various  positions  of  the  given  point. 

6.  To  find  a  point  that  is  \  in.  from  a  given  point  and  equi- 
distant from  two  parallel  lines. 

Discuss  the  problem  for  various  positions  of  the  given  point. 

7.  Find  the  locus  of  the  mid-point  of  a  chord  of  a  given 
length  that  can  be  drawn  in  a  given  circle. 

8.  rind  the  locus  of  the  mid-point  of  a  chord  drawn  through 
a  given  point  within  a  given  circle. 


U-l G 

'0\ 


144  BOOK  II.    PLANE  GEOMETRY 

9.  To  describe  a  circle  that  shall  pass  through  a  given  point 
and  cut  equal  chords  of  a  given  length  from  two  parallels. 

Analysis.   Let  A  be  the  given  point,  BC  and  DE  the  given  parallels, 
MN  the  given  length,  and  O  the  center  of  the  required  circle. 

Since  the  circle  cuts  equal  chords  from  tw^o 

parallels,  what  must  be  the  relative  distance     b 

of  its  center  from  each  ?   Therefore  what  line 

must  be  one  locus  for  0  ?  F 

Draw  the  ±  bisector  of  MN^  cutting  F(?  at  P. 
How,  then,  does  VM  compare  with  the  radius     b~M 
of  the  circle  required  ?   How  shall  we  then  find 
a  point  O  on  FG  that  is  at  a  distance  TM  from  J.  ?    Do  we  then  know 
that  0  is  the  center  of  the  required  circle  ? 

10.  To  describe  a  circle  that  shall  be  tangent  to  each  of  two 
given  intersecting  lines. 

11.  To  find  in  a  given  line  a  point  that  is  equidistant  from 
tAvo  given  points. 

12.  To  find  a  point  that  is  equidistant  from  two  \\  "-^^ 
given  points  and  at  a  given  distance  from  a  third    p    \x'  q 
given  point. 

13.  To  describe  a  circle  that  has  a  given  radius  and  passes 
through  two  given  points. 

14.  To  find  a  point  at  given  distances  from  two  given  points. 

15.  To  describe  a  circle  that  has  its  center  in  a  given  line 
and  passes  through  two  given  points. 

16.  To  find  a  point  that  is  equidistant  from  two  given  points 
and  also  equidistant  from  two  given  intersecting  lines. 

17.  To  find  a  point  that  is  equidistant  from  two  given  points 
and  also  equidistant  from  two  given  parallel  lines.      , 

18.  To  find  a  point  that  is  equidistant  from    c      A'~~^,  ^ 
two  given  intersecting  lines  and  at  a  given  dis- 
tance from  a  given  point. 

19.  To  find  a  point  that  lies  in  one  side  of    '^      '^ 
a  given  triangle  and  is  equidistant  from  the  other  two  sides. 


EXEECISES 


145 


255.  General  Directions  for  solving  Problems.  In  attacking  a 
new  problem  draw  the  most  general  figure  possible  and  the 
solution  may  be  evident  at  once.  If  the  solution  is  not  evident, 
see  if  it  depends  on  finding  a  point,  in  which  case  see  if  two 
loci  can  be  found.  If  this  is  not  the  case,  assume  the  problem 
solved  and  try  to  work  backwards, — the  method  of  analysis. 


EXERCISE  39 

1.  To  draw  a  common  tangent  to  two  given  circles. 

If  the  centers  are  0  and  0'  and  the  radii  r  and  r^,  the  tangent  QR 
seems  to  be  II  to  CKJlf,  a  tangent  from  0'  to  a  circle  whose  radius  is  r  —  r'. 
If  this  is  true,  we  can  easily  reverse  the  process.  Since  there  are  two 
tangents  from  (X,  so  there  are  two  common  tangents. 

In  the  right-hand  figure  tlie  tangent  QR  seems  to  be  II  to  (XM,  a  tangent 
from  CK  to  a  circle  whose  radius  is  r  +  r'.  If  this  is  true,  we  can  easily- 
reverse  the  process.   There  are  four  common  tangents  in  general. 


2.  To  draw  a  common  tangent  to  two  given  circles,  using  the 
following  figures. 


3.  The  locus  of  the  vertex  of  a  right  triangle,  having  a 
given  hypotenuse  as  its  base,  is  the  circle  described  upon  the 
given  hypotenuse  as  a  diameter. 

4.  The  locus  of  the  vertex  of  a  triangle,  having  a  given  base 
and  a  given  angle  at  the  vertex,  is  the  arc  which  forms  with 
the  base  a  segment  in  which  the  given  angle  may  be  inscribed. 


146 


BOOK  II.    PLANE  GEOMETEY 


To  construct  an  isosceles  triangle^  having  given : 

5.  The  base  and  the  angle  at  the  vertex. 

6.  The  base  and  the  radius  of  the  circumscribed  circle. 

7.  The  base  and  the  radius  of  the  inscribed  circle. 

8.  The  perimeter  and  the  altitude.  c- 

Let  ABC  be  the  A  required,  EF  the  given 
perimeter.    The  altitude  CD  passes  through  the  ^^ 


-^-i 


middle  of  EF,  and  the  A  EA  C,  BFC  are  isosceles.  A  D  B 

To  construct  a  right  triangle,  having  given : 
9.  The  hypotenuse  and  one  side. 

10.  One  side  and  the  altitude  upon  the  hypotenuse. 

11.  The  median  and  the  altitude  upon  the  hypotenuse.   - 

12.  The  hypotenuse  and  the  altitude  upon  the  hypotenuse. 

13.  The  radius  of  the  inscribed  circle  and  one  side. 

14.  The  radius  of  the  inscribed  circle  and  an  acute  angle. 

To  construct  a  triangle,  having  given  : 

15.  The  base,  the  altitude,  and  an  angle  at  the  base. 

16.  The  base,  the  altitude,  and  the  angle  at  the  vertex. 

1 7.  One  side,  an  adjacent  angle,  and  the  sum  of  the  other  sides. 

18.  To  construct  an  equilateral  triangle,  hav-  , — .c 
ing  given  the  radius  of  the  circumscribed  circle. 

19.  To  construct  a  rectangle,  having  given  one 
side  and  the  angle  between  the  diagonals. 

20.  Given  two  perpendiculars,  AB  and  CD,  ^ 
intersecting  in  0,  and   a  line  intersecting 
these  perpendiculars  in  E  and  F;   to  con-          ^ 
struct  a  square,  one  of  whose  angles  shall  ^-     q 
coincide  with  one  of  the  right  angles  at  O, 
and  the  vertex  of  the  opposite  angle  of  the 
square  shall  lie  in  EF.    (Two  solutions.) 


A^ 


-.  o 


-r-)iB 


EXERCISES 


147 


/« 


21.  A  straight  rod  moves  so  that  its  ends  con- 
stantly touch  two  fixed  rods  perpendicular  to 
each  other.    Find  the  locus  of  its  mid-point. 

22.  A  line  moves  so  that  it  remains  par- 
allel to  a  given  line,  and  so  that  one  end 
lies  on  a  given  circle.  Find  the  locus  of  the 
other  end. 

23.  Find  the  locus  of  the  mid- 
point of  a  line-segment  that  is  drawn 
from  a  given  external  point  to  a  given 
circle. 

24.  To  draw  lines  from  two  given  points 
P  and  Q  which  shall  meet  on  a  given  line 
AB  and  make  equal  angles  with  AB. 
'.'  Z  BEQ  =  Z  PEC,  .'.  Z  CEP'  =  Z  PEC.    (Why  ?) 

But  it  is  easy  to  make  Z  CEP'=  Z  PEC,  by  mak- 
ing PP'  JlAB,  and  CP'  =  PC,  and  joining  P'  and  Q. 

25.  To  find  the  shortest  path  from  a  point  P  to  a  line  AB 
and  thence  to  a  point  Q.  q 

Prove  that  PE  +  EQ<PF-\-  FQ,  where  Z  BEQ 
=  Z  PEC.  _i 

This  shows  that  a  ray  of  light  from  a  point  to  a  ^  €■ 
plane  mirror  and  thence  to  another  jwint  takes  the  i-'-'' 

shortest  possible  path. 

26.  The  bisectors  of  the  angles  included  by  the  opposite 
sides  (produced)  of  an  inscribed  quadrilateral  intersect  at 
right  angles. 

Arc  ^X—  arc  MD 

=  arc  XB  - 
Arc  YA  -  arc  BN 

=  arc  DY- 
.'.  arc  YX  +  arc  NM 

=  arc  MY  +  arc  XN.   (Why  ?) 
.-.  Z  YIX  =  Z  XIN.    (Why  ?) 
How  does  this  prove  the  proposition  ?   Discuss  the  impossible  case. 


P 


E^-'  F     B 


arc  CM.    (Why  ?) 
arc  NC.   (Why  ?) 


148  BOOK  II.    PLANE  GEOMETRY 

27.  Construct  this  design,  making  the  figure 
twice  this  size. 

Construct  the  equilateral  A.  Then  describe  the  small 
(D  with  half  the  side  of  the  A  as  a  radius.  Then  find 
the  radius  of  the  circumscribing  O. 

28.  A  circular  window  in  a  church  has  a  de- 
sign similar  to  the  accompanying  figure.  Draw 
it,  making  the  figure  twice  this  size. 

This  is  made  from  the  figure  of  the  preceding  exer- 
cise, by  erasing  certain  lines. 

29.  Two  wheels  of  radii  1  ft.  6  in.  and  2  ft.  3  in.  respec- 
tively are  connected  by  a  belt,  drawn  straight  between  the 
points  of  tangency.  The  centers  being  6  ft.  apart,  draw  the 
figure  mathematically.    Use  the  scale  of  1  in.  to  the  foot. 

30.  A  water  wheel  is  broken  and  all  but  a  fragment  is  lost. 
A  workman  wishes  to  restore  the  wheel.    Make 
drawing  showing  how  he  can  construct  a  wheel 
the  size  of  the  original. 

31.  In  this  figure  Zm  =  62°,  ^ndZn=2S°. 
Find  the  number  of  degrees  in  each  of  the 
other  angles,  and  determine  whether  ^^  is 
a  diameter. 

32.  In  this  figure  ZB  =  4.1°,  ZA  =  65°, 
and  ZBDC  =  97°.  Eind  the  number  of 
degrees  in  each  of  the  other  angles,  and 
determine  whether  CD  is  a  diameter. 

33.  Construct  or  explain  why  it  is  im-  ^'         ~ 
possible  to  construct  a  triangle  with  sides  3  in.,  2  in.,  6  in. ; 
also  one  with  sides  5  in.,  7  in.,  12  in.;  also  one 
with  sides  2  in.,  1  in.,  1^  in. 

34.  Show  how  to  draw  a  tangent  to  this  circle  p| 
at  the  point  P,  the  center  of  the  circle  not  being 
accessible. 


.1^^^ 


EXERCISES  149 

EXERCISE  40 

1.  In  a  circle  whose  center  is  O  the  chord  AB  is  drawn  so 
that  Z  BA  O  =  27°.    How  many  degrees  are  there  in  Z.AOB? 

2.  In  a  circle  whose  center  is  0  the  chord  AB  is  drawn  so 
that  Z.BAO  =  25°.  On  the  circle,  and  on  the  same  side  of  AB 
as  the  center  0,  the  point  D  is  taken  and  is  joined  to  A  and  B. 
How  many  degrees  are  there  in  Z.ADB? 

3.  What  is  the  locus  of  the  mid-point  of  a  chord  of  a  circle 
formed  by  secants  drawn  from  a  given  external  point  ? 

4.  In  a  circle  whose  center  is  0  two  perpendiculars  OM  and 
ON  are  drawn  to  the  chords  AB  and  CD  respectively,  and  it 
is  known  that  Z  NMO  =  Z  ONM.   Prove  that  AB=CD. 

5.  Two  circles  intersect  at  the  points  A  and  B.  Through 
A  a  variable  secant  is  drawn,  cutting  the  circles  at  C  and  D. 
Prove  that  the  angle  DBC  is  constant. 

6.  Let  A  and  B  be  two  fixed  points  on  a  given  circle,  and  M 
and  N  be  the  extremities  of  a  rotating  diameter  of  the  same 
circle.  Find  the  locus  of  the  point  of  intersection  of  the 
lines  AM  and  BN. 

7.  Upon  a  line  AB  a,  segment  of  a  circle  containing  240°  is 
constructed,  and  in  the  segment  any  chord  PQ  subtending  an 
arc  of  60°  is  drawn.  Find  the  locus  of  the  point  of  intersection 
of  AP  and  BQ ;  also  of  A  Q  and  BP. 

8.  To  construct  a  square,  given  the  sum  of  the  diagonal  and 
one  side.  ^d 

Let  A  BCD  be  the  square  required,  and  CA  the  di-  j^  ^  A/  \ 
agonal.  Produce  CA,  making  AE  =  AB.  A  ABC  and  """->.,:;,  /' 
ABE  are  isosceles  and  ABAC  =  ZACB  =  i6°.    Find  B 

the  value  of  Z  E.    Construct  Z  CBE.   Now  reverse  the  reasoning. 

The  propositions  in  Exercise  40  are  taken  from  recent  college  entrance 
examination  papers. 


150  BOOK  II.    PLANE  GEOMETEY 

EXERCISE  41 
Review  Questions 

1.  Define  the  word  circle  and  the  principal  terms  used  in 
connection  with  it. 

2.  What  is  meant  by  a  central  angle  ?    How  is  it  measured  ? 

3.  What  is  meant  by  an  inscribed  angle?  How  is  it  measured? 

4.  State  the  general  proposition  covering  all  the  cases  that 
have  been  considered  relating  to  the  measure  of  an  angle  formed 
by  the  intersection  of  two  secants. 

5.  State  all  of  the  facts  you  have  learned  relating  to  equal 
chords  of  a  circle. 

6.  State  all  of  the  facts  you  have  learned  relating  to  unequal 
chords  of  a  circle. 

7.  State  all  of  the  facts  you  have  learned  relating  to  tangents 
to  a  circle. 

8.  How  many  points  are  required  to  determine  a  straight 
line  ?  two  parallel  lines  ?  an  angle  ?  a  circle  ? 

9.  Name  one  kind  of  magnitude  that  you  have  learned  to 
trisect,  and  state  how  you  proceed  to  trisect  this  magnitude. 

10.  In  order  to  construct  a  definite  triangle,  what  parts  must 
be  known  ? 

11.  What  are  the  important  methods  of  attacking  a  new 
problem  in  geometry  ?    Which  is  the  best  method  to  try  first  ? 

12.  What  is  meant  by  determinate,  indeterminate,  and  im- 
possible cases  in  the  solution  of  a  problem  ? 

13.  Distinguish  between  a  constant  and  a  variable,  and  give 
an  illustration  of  each. 

14.  Distinguish  between  inscribed,  circumscribed,  and  escribed 
circles. 

15.  What  is  meant  by  the  statement  that  a  central  angle  is 
measured  by  the  intercepted  arc  ? 


BOOK  III 

PROPORTION.    SIMILAR  POLYGONS 

256.  Proportion.  An  expression  of  equality  between  two 
equal  ratios  is  called  a  proportion. 

257.  Symbols.  A  proportion  is  written  in  one  of  the  fol- 
lowing forms  :  7  =  ~;5   a:b  =  c  :  d\  a  :  b  : :  c  :  d. 

This  proportion  is  read  "  a  is  to  6  as  c  is  to  d  "  ;  or  "  the  ratio  of  a  to  5 
is  equal  to  the  ratio  of  c  to  d." 

258.  Terms.  In  a  proportion  the  four  quantities  compared 
are  called  the  terms.  The  first  and  third  terms  are  called  the 
antecedents;  the  second  and  fourth  terms,  the  consequents. 
The  first  and  fourth  terms  are  called  the  extremes;  the  second 
and  third  terms,  the  means. 

Thus  in  the  proportion  a:b  =  c  :d,  a  and  c  are  the  antecedents,  b 
and  d  the  consequents,  a  and  d  the  extremes,  b  and  c  the  means. 

259.  Fourth  Proportional.  The  fourth  term  of  a  proportion 
is  called  the  fourth  proportional  to  the  terms  taken  in  order. 

Thus  in  the  proportion  a :  6  =  c :  d,  d  is  the  fourth  proportional  to 
a,  &,  and  c. 

260.  Continued  Proportion.  The  quantities  a,  b,  c,  d,  •  •  •  are 
said  to  be  in  continued  proportion,  if  a  :  b  =  b  :  c  =  c  :  d  =  •  -  • . 

If  three  quantities  are  in  continued  proportion,  the  second 
is  called  the  mean  proportional  between  the  other  two,  and  the 
third  is  called  the  third  proportional  to  the  other  two. 

Thus  in  the  proportion  a-.b  =  b:c,b  is  the  mean  proportional  between 
a  and  c,  and  c  is  the  third  proportional  to  a  and  b. 

151 


152  BOOK  III.    PLANE  GEOMETRY 

Proposition  I.    Theorem 

261.  In  any  proportion  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means. 

Given  a:b  =  c:d. 

To  prove  that  ad  =  he. 

Proof.  j  =  ^'  §257 

0      a 

Multiplying  by  bd,  ad  =  bc,  by  Ax.  3. "  q.e.d. 

262.  Corollary  1.  The  mean  proportional  between  tivo 
quantities  is  equal  to  the  square  root  of  their  product. 

For  if  a:h  =  h:c,  then  h^  =  ac  (§  261),  and  h  =  Vac,  by  Ax.  5. 

263.  Corollary  2.  If  the  two  antecedents  of  a  proportion 
are  equals  the  two  consequents  are  equal. 

264.  Corollary  3.  If  the  product  of  two  quantities  is  equal 
to  the  product  of  two  others^  either  two  may  be  made  the  extremes 
of  a  proportion  in  which  the  other  two  are  made  the  means. 

For  if  ad  =  6c,  then,  by  dividing  by  6d,  -  =  -,  by  Ax.  4. 

6      d 

Proposition  II.    Theorem 

265.  If  four  quantities  are  in  proportion,  they  are  in 
proportion  hy  alteimation  ;  that  is,  the  first  term  is  to 
the  third  as  the  second  term  is  to  the  fourth. 


or 


Given 

a:  b  =  c:d. 

To  prove  that 

a:  c  =  b:  d. 

Proof. 

ad  =  be. 

§261 

Dividing  by  cd, 

a      b 
-c  =  d' 

Ax.  4 

a:c  =  b:d,hy  %  257. 

Q.E.D. 

THEORY  OF  PROPORTION  153 

Proposition  III.    Theorem 

266.  If  four  quantities  are  in  proportion,  they  are  in 
proportion  hy  inversion  ;  that  is,  the  second  term  is  to 
the  first  as  the  fourth  term  is  to  the  third. 

Given  a:b  =  c:d. 

To  prove  that  h:  a  =  d:  c. 

Proof.  bc  =  ad.  §261 

Dividing  each  member  of  the  equation  by  ac, 

-  =  ->  Ax.  4 

a      c 

or  b  :  a  =  d  :  c,  hy  ^  257.  Q.  e.  d. 

Proposition  IV.    Theorem 

267.  If  four  quantities  are  in  proportion,  they  are  in 
proportion  hy  composition;  that  is,  the  sum  of  the  first 
two  terms  is  to  the  second  term  as  the  sum  of  the  last 
tivo  terms  is  to  the  fourth  term. 

Given  a:b=c:d. 

To  prove  that  a-\-h  :h  =  c  -\-  d:  d. 

Proof.  j  =  --  §257 

0      d 

Adding  1  to  each  member  of  the  equation, 
a      ^      G      ^ 

a-\-b      c-\-d 
b     ^     d     ' 
.-.  a-[-b:b  =  c  +  d:d,hy  %257.  Q.e.d. 

In  a  similar  manner  it  may  be  shown  that 
a  -{-  b  :  a  =  c  -\-  d  :  c. 


a      ^       c      ^  .       ^ 

-  +  1  =  -+1,  Ax.l 


or 


154  BOOK  III.    PLANE  GEOMETRY 

Proposition  V.    Theorem 

268.  If  four  quantities  are  in  proportion ,  they  are  in 
proportion  hy  division;  that  is,  the  difference  of  the 
first  two  terms  is  to  the  second  term  as  the  difference 
of  the  last  two  terms  is  to  the  fourth  term. 


§257 
Ax.  2 


or 


Given 

a:b  =  c:d. 

To  prove  that 

a  —  b:  h  =  c  —  d:  d. 

Proof. 

a      c 
h~d' 

"h      ^      d      ^' 

a  —  h      c  —  d 

h              d 

.'.a-h:h  =  c-d:d,hj  §257. 

In  a  similar  manner 

it  may  be  shown  that 

a  —  b:a  =  c  —  d:c. 

Q.E.D. 


Proposition  VI.   Theorem 

269.  Li  a  series  of  equal  ratios,  the  sum  of  the  ante- 
cedents is  to  the  su7n  of  the  consequents  as  any  ante- 
cedent is  to  its  consequent. 

Given  a:b  =  c:d=:e:f=g:h. 

To  prove  that     a-\-  e  -\-  e-^  g:h  -\-  d  +/+  h  =  a:h. 

Proof.    Let  r  =  -  =  -  =  - =  'z-- 

0      d     J      I  I 

Then  a  =  hr,     c  =  dr,     e=fr,     g  =  hr.  Ax.  3 

,',a.\-e  +  e-\-g  =  {h  +  d+f+h)r.  Ax.  1 

^   CL -\- c -\-  e -\-  g  a 


Ax.  4 


"b-\-d+f+h  h 

a^c-\-e  +  g:b-\-d  +/+  h  =  a  :  h,  by  §  257.     Q.E.D. 


THEORY  OF  PROPORTION  155 

Proposition  VII.    Theorem 

270.  Like  powers  of  the  terms  of  a  proportion  are  iri 
proportion. 

Given  a:bz=c:  d. 

To  prove  that  a" :  5"  =  ^ :  (^. 

Proof.  X  =  ^-    '  §257 

0      a 

•*•  tt^'t:' by  Ax.  5.  q.e.d. 

Proposition  VIII.    Theorem 

271.  If  three  quantities  are  in  continued  proportion, 
the  first  is  to  the  third  as  the  square  of  the  first  is  to 
the  square  of  the  second. 

Given  a:b=b:c. 

To  prove  that  a:  c  =  a^:b\ 


Proof. 

a^  =  a% 

Iden. 

and 

ae  =  b\ 
a"       a      a^ 
ac       c       b^ 

§261 
Ax.  4 

. • .  a  :c  =  a'^: 

h% 

by  § 

257. 

Q.E.D. 

272.  Nature  of  the  Quantities  in  a  Proportion.  Although  we 
may  have  ratios  of  lines,  or  of  areas,  or  of  solids,  or  of  angles, 
we  treat  all  of  the  terms  of  a  proportion  as  numbers. 

If  b  and  d  are  lines    or   solids,   for  example,  we  cannot 

multiply  each  member  of  -  =  -  by  bd,  as  in  §  261. 

Hence  when  we  speak  of  the  product  of  two  geometric  viagni- 
tudes,  we  mean  the  product  of  the  numbers  that  represent  them 
when  expressed  in  terms  of  a  common  unit. 


166.  BOOK  III.    PLANE  GEOMETEY 

EXERCISE  42 

1.  Prove  that  a:h  =  ma  :  mfib. 

2.  \i.  a\h=^G\d^  and  m  :  71=.])  : q,  prove  that  am :bn=cp  :  dq. 

If  a:b  =  c:  d,  prove  the  following  : 

3.  a:d  =  bc:d\  7.  ma  :  nb  =  me  :  nd. 

4.  l:b  =  c:  ad.  S.  a  —  l:b  =  bc  —  d:bd. 

5.  ad:b  =  c:l.  9.  a-^l:l  =  bc  +  d:d. 

6.  ma  :  b  =  mc  :  d.  10.  1 :  ^c  =  1 :  ad. 

11.  a-\-b:a  —  b  =  c-\-d:c  —  d. 

In  Ex.  11,  use  §  267  and  §  268,  and  Ax.  4.    In  this  case  a,  6,  c,  and  d 
are  said  to  be  in  proportion  by  composition  and  division. 

If  a:b  =  b:  c,  prove  the  following : 

12.  G'.b  =  b'.a.  14.  (^  +  V^) (Z>  -  V^)  =  0. 

13.  a:c  =  P:c''.  15.  ac -l:b -1  =  b -\-l'.l. 
16.  If  2:7=  3: a-,  show  that  2^^  =  21,  andic=10^. 

^mc?  ^Ae  value  of  x  in  the  following  : 

17.  l:7=3:ic.  29.  ic  :  2.7  =  7:  5.4. 

18.  2:9  =  5:ic.  30.  cc  :  8.1  =  0.3  :  0.9. 

19.  4:28  =  3:ic.  31.  2:ic  =  £c:32. 

20.  2:^  =  a;:12.  32.  7:£c  =  cc:28. 

21.  3:5  =  ;:c:9.  33.   l:l-f-x  =  a; -1:  3. 

22.  7:21=£c:5.  34.  5  :£c  -  2  =  a:  +  2  :1. 

23.  3:5  =  £c  4-1:10.  35.  cc2:2a  =  3a:6. 

24.  8:15  =  2ic  +  3:45.  36.  x'.4.a  =  2a^:x\ 

25.  0.8:ic  =  4:9.  37.  a:l  =  x-l:l. 

26.  0.7:a;  =  21:15.  38.  a; +l:ic -1=  3  :  2. 

27.  0.25:cc  =  5:8.  39.  3  :cc +  4  =  a;  -  4:  3. 

28.  a;:1.3  =  4:0.26.  40.  ab:b  =  b-cx'.bc-x. 


PKOPORTIOKAL  LINES  157 

Proposition  IX.    Theorem 

273.  If  a  line  is  drawn  through  two  sides  of  a  tri- 
angle parallel  to  the  third  side,  it  divides  the  tivo  sides 
proportionally. 


Given  the  triangle  ABC,  with  EF  drawn  parallel  to  BC. 

To  prove  that  EB:AE=FC:  AF. 

Case  1.    When  AE  and  EB  are  commensurable. 

Proof.    Assume  that  MB  is  a  common  measure  oi  AE  and  EB. 
Let  MB  be  contained  m  times  in  EB,  and  n  times  in  AE. 

Then  EB:AE  =  m:n. 

{For  m  and  n  are  the  numerical  measures  of  EB  and  AE.) 

At  the  points  of  division  on  EB  and  AE  draw  lines  II  to  BC. 

These  lines  will  divide  A  C  into  m-\-n  equal  parts,  of  which 

FC  will  contain  m  parts,  and  AF  will  contain  n  parts.       §  134 

.•.FC:AF=m:n. 
.'.EB:AE  =  FC:AF,hyAx.S.  Q.e.d. 

For  practical  purposes  this  proves  the  proposition,  for  even  if  AE  and 
EB  are  incommensurable,  we  can,  by  taking  a  unit  of  measure  small 
enough,  find  the  measure  of  AE  and  EB  to  as  close  a  degree  of  approxi- 
mation as  we  may  desire,  just  as  we  can  carry  V2  to  as  many  decimal 
places  as  we  wish,  although  its  exact  value  cannot  be  expressed  rationally. 

On  this  account  many  teachers  omit  the  incommensurable  case  dis- 
cussed on  page  158,  or  merely  require  the  proof  there  given  to  be  read 
aloud  and  explained  by  the  class. 


158  BOOK  III.    PLANE  GEOMETEY 

Case  2.    When  AE  and  EB  are  incommensurable, 
A 


B  (J 

Proof.  Divide  AE  into  a  number  of  equal  parts,  and  apply 
one  of  these  parts  to  EB  as  many  times  as  EB  will  contain  it. 

Since  AE  and  EB  are  incommensurable,  a  certain  number  of 
these  parts  will  extend  from  E  to  some  point  G',  leaving  a 
remainder  GB  less  than  one  of  these  parts. 

Draw  GH  II  to  BC. 

Then  EG:AE  =  FN :  AF.  Case  1 

By  increasing  the  number  of  equal  parts  into  which  AE  is 
divided,  we  can  make  the  length  of  each  part  less  than  any 
assigned  positive  value,  however  small,  but  not  zero. 

Hence  GB,  which  is  less  than  one  of  these  equal  parts,  has 
zero  for  a  limit.  §  204 

And  the  corresponding  segment  HC  has  zero  for  a  limit. 

Therefore        EG  approaches  EB  as  a  limit, 
and  FH  approaches  EC  as  a  limit. 

.*.  the  variable  — —  approaches  -~—  as  a  limit, 

AJii  All/ 

and  the  variable  — —  approaches  — —  as  a  limit. 
AF    ^^  AF 

But  — —  is  always  equal  to  — —  •  Case  1 

.•.f|  =  f|,by.207. 


A 

0 

Fl 

L 

\« 

\ 

\m 

I 

\\ 

PKOPORTIONAL  LINES  159 

274.  Corollary  1.  One  side  of  a  triangle  is  to  either  of 
its  segments  cut  off  hy  a  line  parallel  to  the  base  as  the  third 
side  is  to  its  corresponding  segment. 

For  EB.AE  =  FC.AF.  §273 

By  composition,  EB  +  AE:  AE  =  FC  +  AF:  AF,  §  267 

or  AB:AE  =  AC:AF.  Ax.  11 

275.  Corollary  2.  Three  or  more  parallel  lines  cut  off 
proportional  intercepts  on  any  two  transversals. 

Draw  ^iV  II  to  CD. 
Then  AL  =  CG,  LM  =  GK,  MN  =  KB.         §  127 
Now  AH:AM=  AF:AL=FH:LM,  §274 

and  AH:AM=HB:MN.  §273 

.'.AF'.CG  =FH:GK=HB:KB.  Ax.9        b      N        D 

EXERCISE  43 

1.  In  the  figure  of  §  275,  suppose  AH  =  5  in.,  AF=2  in., 
and  CK=6  in.    Find  the  length  of  CG.  d c 

2.  In  this  square  PQ  is  II  to  AB.   If  a  side  of  the   p 
square  is   10  in.,  Z)^  =  14.14  in.    If  DP  =  3  in., 
what  is  the  length  ot  BQ? 

3.  The  sides  of  a  triangle  are  respectively  3  in.,  4  in.,  and 
5  in.  A  line  is  drawn  parallel  to  the  4-inch  side,  cutting  the 
3-inch  side  1  in.  from  the  vertex  of  the  largest  angle.  Find 
the  length  of  the  two  segments  cut  from  the  longest  side. 

4.  Two  pieces  of  timber  1  ft.  wide  are  fitted  together  at 
right  angles  as  here  shown.  AB  is  8  ft.  long,  AC  6  ft.  long, 
and  the  distance  BCj  along  the  dotted  line, 
is  10  ft.  A  carpenter  finds  it  necessary  to 
saw  along  the  dotted  line.  Find  the  length 
of  the  slanting  cut  across  the  upright  piece ; 
across  the  horizontal  piece. 


160  BOOK  III.    PLANE  GEOMETEY 

Proposition  X.    Theorem 

276.  If  a  line  divides  two  sides  of  a  triangle  jpro^or- 
tionally,  it  is  parallel  to  the  third  side. 


or 


B  G 

Given  the  triangle  ABC  with  EF  drawn  so  that 
EBFC 
AE~  AF' 

To  prove  that  EF  is  II  to  BC. 

Proof.    Suppose  that  EF  is  not  parallel  to  BC. 

Then  from  E  draw  some  other  line,  as  EH,  parallel 

Then  AB  :  AE  =  AC  :  AH. 

{One  side  of  a  A  is  to  either  of  its  segments  cut  off  by  a  line  II 
base  as  the  third  side  is  to  its  corresponding  segment.) 

But  EB:AE  =  FC:AF. 

.-.  EB  +  AE:  AE  =  EC -\-  AFiAF, 

AB:AE  =  AC:AF. 

.'.AC:AF=AC:AH. 

.\AF=AH. 

{For  the  two  antecedents  are  equal.) 

.'.  EF  and  EH  must  coincide. 

{For  their  end  points  coincide.) 

EH  is  II  to  BC. 

.'.  EF,  which  coincides  with  EH,  is  II  to  BC. 

This  proposition  is  the  converse  of  Prop.  IX. 


But 


to^C. 

§274 
to  the 

Given 
§267 

Ax.  11 
Ax.  8 
§263 

Post.  1 

Const. 

Q.E.D. 


PKOPOETIONAL  LINES  161 

277.  Dividing  a  Line  into  Segments.  If  a  given  line  AB  is 
divided  at  P,  a  point  between  the  extremities  A  and  B,  it  is 
said  to  be  divided  internally  into  the  segments  AP  and  PB'j 
and  if  it  is  divided  at  P',  a  point  in  the  prolongation  of  BA,  it  is 
said  to  be  divided  externally  into  the  segments  AP'  and  P'B. 


P'  A  P  B 

In  either  case  the  length  of  the  segment  is  the  distance  from  the 
point  of  division  to  an  extremity  of  the  line.  If  the  line  is  divided 
internally,  the  sum  of  the  segments  is  equal  to  the  line  ;  and  if  the  line 
is  divided  externally,  the  difference  of  the  segments  is  equal  to  the  line. 

Suppose  it  is  required  to  divide  the  given  line  AB  internally 
and  externally  in  the  same  ratio ;  as,  for  example,  in  the  ratio 
of  the  two  numbers  3  and  5. 


P'  APE 

We  divide  AB  into  3  +  5,  or  8,  equal  parts,  and  take  3  parts 
from  A  ;  we  then  have  the  point  P,  such  that 

AP'.PB  =  Z:n.  (1) 

Secondly,  we  divide  AB  into  5  —  3,  or  2,  equal  parts,  and  lay 
off  on  the  prolongation  of  BA  three  of  these  equal  parts ;  we 
then  have  the  point  P',  such  that 

ylP':P'5  =  3:5.  (2) 

Comparing  (1)  and  (2),  we  have 

AP:PB  =  AP''.P'B. 

278.  Harmonic  Division.  If  a  given  straight  line  is  divided 
internally  and  externally  into  segments  having  the  same  ratio, 
the  line  is  said  to  be  divided  harmonically. 

Thus  the  line  AB  has  just  been  divided  internally  and  externally  in 
the  same  ratio,  .3  :  5,  and  ^  JB  is  therefore  said  to  be  divided  harmonically 
at  P  and  P'  in  the  ratio  3:5. 


162  BOOK  III.    PLANE  GEOMETRY 

Proposition  XI.    Theorem 

279.  The  bisector  of  an  angle  of  a  triangle  divides 
the  opposite  side  into  segmeiits  which  are  proportional 
to  the  adjacent  sides. 


'--^G 


Given  the  bisector  of  the  angle  C  of  the  triangle  ABC^  meeting 
ABatM, 

To  prove  that  AM:  MB  =  CA:  CB. 

Proof.    Erom  A  draw  a  line  II  to  MC. 

This  line  must  meet  BC  produced,  because  BC  and  MC 
cannot  both  be  parallel  to  the  same  line.  §  94 

Let  this  line  meet  BC  produced  at  E. 
Then  AM:MB=EC  :CB.  §273 

{If  a  line  is  drawn  through  two  sides  of  a  A  parallel  to  the  third  side,  it 
divides  the  two  sides  proportionally.) 

Also  ZACM=ZCAE,  §  100 

{Alt.-int.  A  of  II  lines  are  equal.) 

and  Z  MCB  =  ZAEC.  §  102 

{Ext.-int.  A  of  II  lines  are  equal.) 

But  ZACM=ZMCB.  Given 

,'.ZCAE  =  ZAEC.  Ax.  8 

.\EC=CA.  §76 

Put  CA  for  its  equal  ^C  in  the  first  proportion. 

Then  AM:MB=  CA  :CB,hy  Ax.  9.  Q. E. D. 


PEOPOETIONAL  LINES  163 

Pboposition  XII.    Theorem 

280.  The  bisector  of  an  exterior  angle  of  a  triangle 
divides  the  opposite  side  externally  into  segments  ivhich 
are  proportional  to  the  adjacent  sides. 


A  ^ 

Given  the  bisector  of  the  exterior  angle  ECA  of  the  triangle  ABCy 
meeting  BA  produced  at  M'. 

To  prove  that       AM' :  M'B  =  CA:CB. 

Proof.    Draw  AF  II  to  M'C,  meeting  EC  at  F. 

Then  AM':  M'B=FC -.CB.  §274 

{One  side  of  a  A  is  to  either  of  its  segments  cut  off  by  a  line  II  to 
the  base  as  the  third  side  is  to  its  corresponding  segment.) 

Now                           Z  ECM'=  Z  CFA,  §  102 

and                                ZM'CA=ZFAC.  §100 

But                             ZECM'=ZM'CA.  Given 

.'.ZCFA=ZFAC.  Ax.  8 

.-.  CA  =  FC.  §  76 

Put  CA  for  its  equal  FC  in  the  first  proportion. 

Then                           AM':  M'B  =  CA :  CB,  by  Ax.  9.  Q.e.d. 

Discussion.   In  case  CA  =  CB,  what  is  the  arrangement  of  the  lines  ? 

281.  Corollary.  The  bisectors  of  the  interior  angle  arid  the 
exterior  angle  at  the  same  vertex  of  a  triangle,  meeting  the 
opposite  side,  divide  that  side  harmonically. 


164  BOOK  III.    PLANE  GEOMETRY 

EXERCISE  44 

1.  In  a  triangle  ABC,  AB  =  6.5,  CA  =  6,BC  =  7.  Find  the 
segments  oi  AB  made  by  the  bisector  of  the  angle  C 

2.  In  a  triangle  ABC,  CA  =  7.5,  BC  =  7,AB  =  8.  Find  the 
segments  of  CA  made  by  the  bisector  of  the  angle  B. 

3.  The  sides  of  a  triangle  are  12, 16,  20.  Find  the  segments 
of  the  sides  made  by  bisecting  the  angles. 

4.  If  a  spider,  in  making  its  web, 
makes  A'B'  II  to  AB,  B'C  II  to  BC,  C'D' 
II  to  CD,  D'E'  II  to  DE,  and  E'F'  II  to  EF, 
and  then  runs  a  line  from  F'  II  to  FA, 
will  it  strike  the  point  A '  ?    Prove  it. 

5.  From  any  point  O  within  the  triangle  ABC  the  lines 
OA,  OB,  OC  are  drawn  and  are  bisected  respectively  by  A',  B', 
and  C.    Prove  that  Z CBA  =  Z C'B'A'. 

6.  Prove  Ex.  5  if  the  point  0  is  outside  the  triangle. 

7.  From  any  point  0  within  the  quadrilateral  ABCD  lines  are 
drawn  to  the  vertices  A,B,C,  D,  and  are  bisected  by  A',B',C',  D'. 
Prove  that  Z  CBA  =  Z  C'B'A '. 

8.  If  a  pendulum  swings  at  the  point  0,  cutting  two  paral- 
lel lines  at  P  and  Q  respectively,  the  ratio  OP :  OQ  is  constant. 

9.  Through  a  fixed  point  P  a  line  is  drawn  cutting  a  fixed 
line  at  X.  PX  is  then  divided  at  Y  so  that  the  ratio  PYiYX 
is  constant.  Find  the  locus  of  the  point  F  as  Z  moves  along 
the  fixed  line. 

10.  From  the  point  P  on  the  side  CA  of  the  triangle  ABC 
parallels  to  the  other  sides  are  drawn  meeting  ABin  Q  and  BC 
in  R.    Prove  that  AQ:  QB  =  BR  :  RC. 

11.  In  the  triangle  ABC,  P  and  Q  are  taken  on  the  sides  CA 
and  BC  so  that  AP :  PC  =  BQ:  QC.  AR  is  then  drawn  parallel 
to  PB,  meeting  CB  produced  in  R.  Prove  that  CB  is  the  mean 
proportional  between  CQ  and  CR. 


SIMILAR  POLYGONS 


165 


282.  Similar  Polygons.  Polygons  that  have  their  correspond- 
ing angles  equal,  and  their  corresponding  sides  proportional, 
are  called  similar  polygons. 


Thus  the  polygons  ABODE  and  A'B'C'iyE'  are  similar,  if  the  A  A,  B, 
C,  D,  E  are  equal  respectively  to  the  A  A%  B%  C",  i/,  E\  and  if 
AB  :  A'B'  =  BC  :  B'C  =  CD  :  C'XK  =  DE  :  D'E'  =  EA  :  E'A\ 
Similar  polygons  are  commonly  said  to  be  of  the  same  shape. 

283.  Corresponding  Lines.  In  similar  polygons  those  lines 
that  are  similarly  situated  with  respect  to  the  equal  angles 
are  called  corresponding  lines. 

Corresponding  lines  are  also  called  homologous  lines. 

284.  Ratio  of  Similitude.  The  ratio  of  any  two  corresponding 
lines  in  similar  polygons  is  called  the  ratio  of  similitude  of 
the  polygons. 

The  primary  idea  of  similarity  is  likeness  of  form.  The  two 
conditions  necessary  for  similarity  are : 

1.  For  every  angle  in  one  of  the  figures  there  must  he  an 
equal  angle  in  the  other. 

2.  The  corresponding  sides  must  he  proportional. 

Thus  Q  and  Q!  are  not  similar ;  the  corresponding  sides  are 
proportional,  but  the  corresponding  angles  are  not  equal.  Also 
R  and  it'  are  not  similar;  the  corresponding  angles  are  equal, 
but  the  corresponding  sides  are  not  proportional. 


Q' 


In  the  case  of  triangles  either  condition  implies  the  other. 


166  BOOK  III.    PLANE  GEOMETRY 

Proposition  XIII.    Theorem 
285.  Two  rnutually  equiangular  triangles  are  similar. 


Given  the  triangles  ABC  and  A'B^C\  having  the  angles  A,  B^  C 
equal  to  the  angles  A',  B\  C  respectively. 

To  prove  that  the  A  ABC  and  A'B'C'  are  similar.. 
Proof.    Since  the  A  are  mutually  equiangular,  Given 

we  have  only  to  prove  that 
AB  :  A^B'  =  AC  :  A'C  =  BC  :  B'C.  §  282 

Place  the  A^'^'C*'  on  the  A^^C  so  thatZC  shall  coincide 
with  its  equal,  the  Z  C,  and  A'B'  take  the  position  PQ.    Post.  5 

Then                                 Aj^  =  ^  A.  Given 

.-.  PQ  is  II  to  AB.  §  103 

.'.AC:PC  =  BC:  QC ;  §274 

that  is,                         AC  :A'C'  =  BC  iB'C.  Ax.  9 

Similarly,  by  placing  the  A  A'B'C  on  the  A  ABC  so  that  Z  B' 
shall  coincide  with  its  equal,  the  Z  B,  we  can  prove  that 
AB:A'B'  =  BC:B'C'. 
.\  AB  :  A'B'  =  AC  :  A'C  =  BC  :  B'C.      Ax.  8 
.-.  A  ABC  is  similar  to  A  A'B'C,  by  §  282.         q.e.d. 

286.  Corollary  1.   Two  triangles  are  similar  if  two  angles 
of  the  one  are  equal  respectively  to  two  angles  of  the  other. 

287.  Corollary  2.   Two  right  triangles  are  similar  if  an 
acute  angle  of  the  one  is  equal  to  an  acute  angle  of  the  other. 


SIMILAR  POLYGONS 


167 


Proposition  XIV.    Theorem 

288.  If  two  triangles  have  an  angle  of  the  one  equal 
to  an  angle  of  the  other ^  and  the  including  sides  propor- 
tional, they  are  similar. 


Given  the  triangles  ABC  and  A'B'C\  with  the  angle  C  equal  to 
the  angle  C  and  with  CA  :  C'A'  =  CB :  C^B\ 

To  prove  that  the  A  ABC  a7id  A'B'C  are  similar. 
Proof.    Place  the  A  A'B'C  on  the  A  ABC  so  that  Z.C'  shall 
coincide  with  its  equal,  the  Z  C.  Post.  5 

Then  the  A  A'B'C  will  take  the  position  of  the  A  PQC. 
CA  _  CB  , 
CA'~C^'' 
CA       CB 
CP~  CQ 
CA-CP      CB  —  CQ 


Now 


that  is, 


Given 


Ax.  9 


or 


CP 
PA 
CP 

..'.  PQ  is 


CQ 
QB 
CQ 
to  AB. 


{If  a  line  divides  two  sides  of  a  A  proportionally,  it  is 
.'.  Ap  =z  Z.A,  and  Aq  =  Z.B. 
Now  /LC  =  AC'. 

.'.  A  PQC  is  similar  to  A  ABC. 
.'.  A  A'B'C  is  similar  to  A  ABC. 


268 


§276 
to  the  third  side.) 

§102 

Given 

§285 


Q.  E.  D. 


168 


BOOK  III.    PLANE  GEOMETRY 


Proposition  XY.    Theorem 

289.  If  two  triangles  have  their  sides  respectively 
proportional^  they  are  similar. 


A  B  A'  B' 

Given  the  triangles  ABC  and  A'B'C\  having 

AB :  A'B'  =BC:B'C'=CA:  CA  \ 

To  prove  that  the  A  ABC  and  A'B'C  are  similar. 

Proof.    Upon  CA  take  CP  equal  to  CA\  and  upon  CB  take 
CQ  equal  to  C'^';  and  draw  PQ. 


Now 
Or,  since 

Also 


CA  :  C'A'  =  BC:B'C'. 

CP  =  CA',  and  CQ  =  C'B', 

CA:  CP  =  CB:CQ. 

ZC  =  ZC. 

A  ABC  and,PQC  are  similar. 


Given 

Const. 

Ax.  9 

Iden. 

§288 


{If  two  A  have  an  angle  of  the  one  equal  to  an  angle  of  the  other,  and 
the  including  sides  proportional,  they  are  similar.) 


that  is, 
But 


But 


.\CA:CP  =  AB:PQ; 

CA  :  CA'  =  AB  :  PQ. 

CA:CA'  =  AB:A'B'. 
.-.  AB:PQ  =  AB:A'B'. 
.\PQ  =  A'B'. 

Hence  the  A  PQC  and  A'B'C  Sive  congruent. 

APQC  has  been  proved  similar  to  A  ABC. 

.'.A  A'B'C  is  similar  to  A  ABC 


§282 
Ax.  9 
Given 
Ax.  8 
§263 

§80 

Q.E.D. 


SIMILAR  POLYGOJ^S 
Proposition  XVI.    Theorem 


169 


290.  TiDo  triangles  ivhich  have  their  sides  respectively 
jjarallel,  or  ^respectively  perpendicular,  are  similar. 


Given  the  triangles  ABC  and  A'B'C'j  with  their  sides  respec- 
tively parallel ;  and  the  triangles  DEF  and  D'E'F'y  with  their  sides 
respectively  perpendicular. 

To  prove  that      1.   the  A  ABC  and  A'B'C  are  similar  ; 
2.   the  A  DEF  and  D'E'F'  are  similar. 

Proof.    1.  Produce  EC  and  AC  to  B'A',  forming  A  x  and  y. 
Then  Z.B  =  Zx(^  100),  Siud  ZB' =  Zx.  §102 

.\ZB  =  ZB'.  Ax.  8 

In  like  manner,  ZA  =  ZA'. 

.-.  A  ABC  is  similar  to  A  A'B'C.  §  286 

2.  Produce  DE  and  FD  to  meet  D'E'  and  F'D'  at  P  and  R. 

The  quadrilateral  E'QEP  has  zip  and  q  right  angles.  Given 

.*.  A  E'  and  PEQ  are  supplementary.  §  144 

But                 A  y  and  Pi?Q  are  supplementary.  §  43 

Therefore                         Zy  =  ZE'.  §  58 

In  like  manner,  Zx  =  ZD'. 

.-.A  X>£:f  is  similar  to  A  D'E'F',  by  §  286.  Q.e.d. 

Discussion.  The  parallel  sides  and  the  perpendicular  sides  respectively 
are  corresponding  sides  of  the  triangles. 


170  BOOK  III.    PLANE  GEOMETRY 

Pkoposition  XVII.    Theorem 

291.  The  perimeters  of  two  similar  polygons  have  the 
same  7'atio  as  any  two  corresponding  sides. 

D 


Given  the  two  similar  polygons  ABCDE  and  A'B'C'D'E',  with 
p  and  />'  representing  their  respective  perimeters. 

To  prove  that  p  ip'  =  AB  :  A'B'. 

AB  _  BC_  _   CD_  _  DE  _  EA_ 
A'B'  ~  B'C  ~  CD'  ~  D'E'  ~  EA''  ^ 


AB  -{-  BC  -{-  CD  -\-  DE  -{-  EA      _  AB 
A 'B'  +  B'C  +  C"X>'+  D'E'  +  E'A'~  A 'B'' 


§269 


r,2):p'  =  AB:A'B'jhyAx.9.  q.e.d. 

EXERCISE  45 

1.  The  corresponding  altitudes  of  two  similar  triangles  have 
the  same  ratio  as  any  two  corresponding  sides. 

2.  The  base  and  altitude  of  a  triangle  are  15  in.  and  7  in. 
respectively.  The  corresponding  base  of  a  similar  triangle  is 
3.75  in.    Find  the  corresponding  altitude. 

3.  If  two  parallels  are  cut  by  three  concurrent  transversals, 
the  corresponding  segments  of  the  parallels  are  proportional. 

4.  The  point  P  is  any  point  on  the  side  OX  of  the  angle 
XOY.  From  P  a  perpendicular  PQ  is  let  fall  on  OY.  Prove 
that  for  any  position  of  P  on  OX  the  ratio  OP  :  PQ  is  constant, 
and  the  ratio  PQ:  OQ  is  constant. 


SIMILAR  POLYGONS 


171 


5.  In  drawing  a  map  of  a  triangular  field  with  sides  75  rd., 
60  rd.,  and  50  rd.  respectively,  the  longest  side  is  drawn  1  in. 
long.    How  long  are  the  other  two  sides  drawn  ? 

6.  This  figure  represents  part  of  a  diagonal  scale  used  by 
draftsmen.  The  distance  from  0  to  10 
is  1  centimeter,  or  10  millimeters.  Show 
how  to  measure  5  mm, ;  1  mm. ;  0.9  mm. ; 
0.5  mm. ;  1.5  mm.  On  what  proposition 
does  this  depend  ?  _.     _     . 

7.  This  figure  represents  a  pair  of  proportional 
compasses  used  by  draftsmen.  By  adjusting  the 
screw  at  0,  the  lengths  OA  and  OC,  and  the  corre- 
sponding lengths  OB  and  OD,  may  be  varied  pro- 
portionally. Prove  that  AOAB  is  always  similar 
to  AOCD.  If  0^  =  3  in.  and  OC  =  5  in.,  then  AB 
is  what  part  of  CD? 

8.  ABCD  is  any  polygon  and  P  is  any 
point.  On  ^  P  any  point  A '  is  taken  and  A  'B' 
is  drawn  parallel  to  AB  as  shown.  Then 
B'C'&nd  CD'  are  drawn  parallel  to  BC  and 
CD.  Is  D'A'  parallel  to  DA  ?  Is  A'B'C'D' 
similar  to  ABCD?    Prove  it. 

9.  If  two  circles  are  tangent  externally,  the  corresponding 
segments  of  two  lines  drawn  through  the  point  of  contact  and 
terminated  by  the  circles  are  proportional. 

10.  If  two  circles  are  tangent  externally,  their  common  ex- 
ternal tangent  is  the  mean  proportional  between  their  diameters. 

11.  AB  and  AC  are  chords  drawn  from  any  point  /I  on 
a  circle,  and  AD  is  ^  diameter.  If  the  tangent  at  D  intersects 
AB  and  A  C  at  i?  and  F,  the  triangles  ABC  and  A  EF  are  similar. 

12.'  If  AD  and  BE  are  two  altitudes  of  the  triangle  ABC^  the 
triangles  DEC  and  ABC  are  similar. 


p^-: 


172 


BOOK  III.    PLANE  GEOMETRY 


Proposition  XVIII.    Theorem 

292.  If  tivo  polygons  are  similar,  they  can  he  sepa- 
rated hito  the  same  number  of  triangles,  similar  each  to 
each,  and  similarly  placed. 


Given  two  similar  polygons  ABCDE  and  A'B'CD'E^  with  angles 
A,  B,  C,  2>,  E  equal  to  angles  ^',  5',  C,  D\  E'  respectively. 

To  prove  that  ABODE  and  A'B'C'D'E'  can  be  separated 
into  the  same  number  of  triangles,  similar  each  to  each,  and 
similarly  placed. 

Proof.  Draw   the    corresponding    diagonals    DA,   D'A',    and 


DB,  D'B'. 

Since  Z.E  =  Z.E\ 

§282 

and 

DE:D'E'  =  EA  :  E'A', 

§282 

.'.  ADEA  and  D'E'A'  are  similar. 

§288 

In  like  manner, 

ADBC  and  D'B'C  are  similar. 

Purthermore 

ZBAE  =  ZB'A'E', 

§282 

and 

ZDAE  =  ZDA'E'. 

§282 

By  subtracting, 

ZBAD  =  ZBA'D'. 

Ax.  2 

Now 

DA  :DA'  =  EA  :  E'A', 

§282 

and 

AB:A'B'  =  EA  :  E'A'. 

§282 

. 

'.  DA:D'A'  =  AB:A'B'. 

Ax.  8 

.-.A  DAB  and  D'A'B'  are  similar. 

by  §  288. 

Q.E.D. 

SIMILAR  POLYGONS 


173 


Proposition  XIX.    Theorem 

293.  If  tivo  polygons  are  composed  of  the  same  num- 
ber of  triangles,  similar  each  to  each,  and  similarly 
pjlaced,  the  polygons  are  similar. 


A  B  A'  B' 

Given  two  polygons  ABCDE  and  A'B^C'D'E'  composed  of  the 
triangles  DEA^  DAB^  DBC,  similar  respectively  to  the  triangles 
D'E'A\  D'A'B',  D'B'C,  and  similarly  placed. 

To  prove  that  ABCDE  is  mnilar  to  A'B'C'D'U'. 

Proof.  ZE  =  ZE'. 

Also  ZDAE  =  Z  D'A  'E'j 

and  ZBAD  =  ZB'A'D'. 

By  adding,  ZBAE  =  ZB'A 'E'. 

Similarly    Z  CBA  =  Z  C'B'A ',  and  Z  EDC  =  Z  E'D'C. 

Again,  ZC  =  ZC'.  §  282 

Hence  the  polygons  are  mutually  equiangular. 

DE  _  EA         DA        AB        DB        BC         CD 
~  E'A' 


§282 

§282 
Ax.  1 


Also 


282 


D'E'      E'A'      D'A'      A'B'      D'B'      B'C      CD' 

Hence  the  polygons  are  not  only  mutually  equiangular  but 
they  have  their  corresponding  sides  proportional. 

Therefore  the  polygons  are  similar,  by  §  282.  q.e.d. 

This  proposition  is  the  converse  of  Prop.  XVIII. 


174  BOOK  III.    PLANE  GEOMETRY 

Proposition  XX.    Theorem 

294.  If  m  a  right  triangle  a  perpeiidicular  is  clraivn 
from  the  vertex  of  the  right  aiigle  to  the  hypotenuse : 

1.  The  triangles  thus  formed  are  similar  to  the  given 
triangle,  and  are  similar  to  each  other. 

2.  The  perpendicidar  is  the  mean  proportional  he- 
tween  the  segments  of  the  hypotenuse. 

3.  Each  of  the  other  sides  is  the  mean  proportional 
hetiveen  the  hypotenuse  and  the  segment  of  the  hypote- 
nuse adjacent  to  that  side, 

c 


A  F  B 

Given  the  right  triangle  ABC^  with  CF  drawn  from  the  vertex 
of  the  right  angle  C,  perpendicular  to  AB. 

1.  To  prove  that  the  A  BCA,  CFA,  BFC  are  similar. 
Proof.    Since  the  Z.  a^  is  common  to  the  rt.  A  CFA  and  BCA, 

.'.  these  A  are  similar.  §287 

Since  the  Z  Z>  is  common  to  the  rt.  A  BFC  and  BCA, 

.'.  these  A  are  similar.  §  287 

Since  the  A  CFA  and  BFC  are  each  similar  to  A  BCA, 

.'.  these  A  are  mutually  equiangular,  §  282 

Therefore  the  A  CFA  and  BFC  are  similar,  by  §  285.     Q.  e.  d. 

2.  To  prove  that     AF  :CF=CF:  FB. 
Proof.    In  the  similar  A  CFA  and  BFC, 

AF:  CF=CF:FB,  by  §  282.  q.e.d. 


NUMEKICAL  PROPERTIES  OF  FIGURES      175 

3.    To  prove  that     AB:AC  =  AC:AF, 
and  AB:BC  =  BC:BF. 

Proof.    In  the  similar  A  EC  A  and  CFA, 

AB:AC  =  AC:AF.  §282 

In  the  similar  ABC  A  and  BFC, 

AB:BC  =  BC:  BF,  by  §  282.  q.e.d. 

295.  Corollary  1.  The  squares  on  the  two  sides  of  a  right 
triangle  are  proportional  to  the  segments  of  the  hypotenuse 
adjacent  to  those  sides.  , 

From  the  proportions  in  §  294,  3, 

AAJ^  =  ABx  AF,  and  liC^  =  AB  x  BF.  §  261 

Hence  -^^AB^AF^A^  ^^  , 

BC^       AB  X  BF      BF 

296.  Corollary  2.  The  square  on  the  hypotenuse  and  the 
squai^e  on  either  side  of  a  right  triangle  are  proportional  to 
the  hypotenuse  and  the  seg7nent  of  the  hypotenuse  adjacent 
to  that  side. 

Since                                 AB^  ^  AB  x  AB,  Iden. 

and,  as  in  §  295,                  AC^  =  AB  x  AF,  §  261 

AB^  _  ABx  AB  _AB  ^^  ^ 
''  AC''~  ABx  AF~AF' 

297.  Corollary  3.  The  perpendicular  from  any  point  on 
a  circle  to  a  diameter  is  the  mean  proportional  hehveen  the 
segments  of  the  diameter.  g^ 

298.  Corollary  A.  If  a  perpendicular 
is  drawn  from  any  point  on  a  circle  to  a 
diameter^   the   chord  from    that  point    to   ^ 
either  extremity  of  the  diameter  is  the  mean  proportional  he- 
tiveen  the  diameter  and  the  segment  adjacent  to  that  chord. 


176  BOOK  III.    PLANE  GEOMETRY 

EXERCISE  46 

1.  The  perimeters  of  two  similar  polygons  are  18  in.  and 
14  in.  If  a  side  of  the  first  is  3  in.,  find  the  corresponding  side 
of  the  second. 

2.  In  two  similar  triangles,  ABC  and  A'B'C',  AB  =  6  in., 
BC  =  7  in.,  CA  =  S  in.,  and  A'B'  =  9  in.    Eind  B'C  and  C'A'. 

3.  The  corresponding  bases  of  two  similar  triangles  are 
11  in.  and  13  in.  The  altitude  of  the  first  is  6  in.  Find  the 
corresponding  altitude  of  the  second. 

4.  The  perimeter  of  an  equilateral  triangle  is  51  in.  Eind 
the  side  of  an  equilateral  triangle  of  half  the  altitude. 

5.  The  sides  of  a  polygon  are  2  in.,  2^  in.,  3|  in.,  3  in.,  and 
5  in.  Eind  the  perimeter  of  a  similar  polygon  whose  longest 
side  is  7  in. 

6.  The  perimeter  of  an  isosceles  triangle  is  13,  and  the  ratio 
of  one  of  the  equal  sides  to  the  base  is  1§.    Eind  the  three  sides. 

7.  The  perimeter  of  a  rectangle  is  48  in.,  and  the  ratio  of 
two  of  the  sides  is  f .    Eind  the  sides. 

8.  In  drawing  a  map  to  the  scale  y^^oVoo)  what  length  will 
represent  the  sides  of  a  county  that  is  a  rectangle  25  mi.  long 
and  10  mi.  wide  ?    Answer  to  the  nearest  tenth  of  an  inch. 

9.  Two  circles  touch  at  P.  Through  P  three  lines  are 
drawn,  meeting  one  circle  in  A,  B,  C,  and  the  other  in 
A\  B',  C  respectively.  Prove  the  triangles  ABC,  A'B'C'  similar. 

10.  If  two  circles  are  tangent  internally,  all  chords  of  the 
greater  circle  drawn  from  the  point  of  contact  are  divided  pro- 
portionally by  the  smaller  circle. 

11.  In  an  inscribed  quadrilateral  the  product  of  the  diagonals 
is  equal  to  the  sum  of  the  products  of  the  opposite 
sides. 

Draw  D^,  making  ZEDC  =  Z  ABB.   The  A  ABB  and 
ECB  are  similar ;  and  the  A  BCB  and  AEB  are  similar. 


NUMERICAL  PROPERTIES  OF  FIGURES      177 

Proposition  XXI.    Theorem 

299.  If  tivo  chords  intersect  ivithin  a  circle,  the  prod- 
uct of  the  segments  of  the  one  is  equal  to  the  product 
of  the  segments  of  the  other. 


Given  the  chords  AB  and  CDy  intersecting  at  P. 
To  prove  that        PAxFB=FCx  PD. 
Proof.  Draw  A  C  and  BD. 

Then  since  Aa  =  Z.a\  §  214 

(Each  is  measured  by  |  arc  CB.) 

and  Zc  =  Zc\  §214 

{Each  is  measured  by  |  arc  DA.) 

.-.  the  A  CPA  and  BPD  are  similar.  §  286 

.-.  PA  :PD=PC:PB.  §282 

.-.  PAxPB=PC  xPD,hj  ^261.  Q.E.D. 

300.  Corollary.   If  two  chords  intersect  within  a  circle, 

the  segments  of  the  one  are  reciprocally  proportional  to  the 

segments  of  the  other. 

This  means,  for  example,  that  PA  :  PD  equals  the  reciprocal  of  PB :  PC, 
or  equals  PC  :  PB,  as  shown  above. 

301.  Secant  to  a  Circle.  A  secant  from  an  external  j^oint  to  a 
circle  is  understood  to  mean  the  segment  of  the  secant  that  lies 
between  the  given  external  point  and  the  second  j^oint  of  inter- 
section of  the  secant  and  circle. 


178  BOOK  III.    PLANE  GEOMETRY 

Proposition  XXII.    Theorem 

302.  If  from  a  point  outside  a  circle  a  secant  and  a 
tangent  are  draion,  the  tangent  is  the  7nean  propor- 
tional hetiveen  the  secant  and  its  external  segment. 


Given  a  tangent  AD  and  a  secant  AC  drawn  from  the  point  A  to 
the  circle  BCD. 

To  prove  that         AC.  AD  =  AD:  AB. 
Proof.  Draw  DC  and  DB. 

Now  Z  c  is  measured  by  ^  arc  DB,  §  214 

and  Z  c'  is  measured  by  i  arc  DB.  §  220 

.'.Z.c  =  Z.c\ 
Then  in  the  A  ADC  and  ABD, 

Z.a  =  /.a,  Iden. 

and  Ac  —  Ac\ 

.-.  A^Z>C  and  ^5/)are  similar.  §286 

.•.AC\AD=.AD'.AB,  by  §282.  q.e.d. 

303.  Corollary.  If  from  a  fixed  'point  outside  a  circle  a 
secant  is  dratvn,  the  product  of  the  secant  and  its  external 
segment  is  constant  in  whatever  direction  the  secant  is  drawn. 

Smce  AC:AD  =  AD:AB,  §302 

.-.  AC  X  AB  =  AD^  §261 

Since  AD  is  constant  (§  192),  therefore  AC  x  AB  is  constant. 


NUMERICAL  PROPERTIES  OF  FIGURES      179 

Proposition  XXIII.    Theorem 

304.  The  square  on  the  bisector  of  an  angle  of  a  tri- 
angle is  equal  to  the  product  of  the  sides  of  this  angle 
diminished  hy  the  product  of  the  segments  made  by  the 
bisector  upon  the  third  side  of  the  triangle. 


~--C 


D 
Given  the  line  CP  bisecting  the  angle  ACB  of  the  triangle  ABC. 
To  prove  that     CP''  =  CA  x  BC-AP  x  PB. 
Proof.    Circumscribe  the  OBCA  about  the  A  ABC.         §  240 
Produce  CP  to  meet  the  circle  in  Z>,  and  draw  BD. 
Then  in  the  A  BCD  and  PC  A, 


and 


/.m  =  Z.  m\ 

Given 

Aa'  =  A  a. 

§214 

{Each  is  measured  by  I  arc  BC.) 

•.  the  A  BCD  and  PC  A  are  similar. 

§286 

.-.  CD:CA=BC:  CP. 

§282 

.-.  CAxBC=CDX  CP 

§261 

=  {CP-\-PD)CP 

Ax.  9 

=  CP''  +  CP  X  PD. 

CPXPD=APXPB. 

§299 

.-.  CA  xBC=CP^-{-APxPB. 

Ax.  9 

CP^  =CAxBC  -APx  PB,  by  Ax.  2. 

Q.E.D. 

But 


This  theorem  enables  us  to  compute  the  bisectors  of  the  angles  of  a 
triangle  terminated  by  the  opposite  sides,  if  the  sides  are  known.  The 
theorem  may  be  omitted  without  destroying  the  sequence. 


180  BOOK  III.    PLANE  GEOMETRY 

Proposition  XXIV.    Theorem 

305.  In  any  triangle  tlie  jproduct  of  two  sides  is  equal 
to  the  product  of  the  diameter  of  the  circumscribed  circle 
hy  the  altitude  upon  the  third  side. 


Given  the  triangle  ABC  with  CP  the  altitude,  ADBC  the  circle 
circumscribed  about  the  triangle  ABC^  and  CD  a  diameter. 

To  prove  that        CAxBC=CDx  CP, 

Proof.  Draw  BD. 

Then  in  the  A  APC  and  DBC, 

Z  CPA  is  a  rt.  Z,  Given 

Z  CBD  is  a  rt.  Z,  §  215 

Z  <t  is  measured  by  \  arc  BC, 

and  Z  a'  is  measured  by  \  arc  BC.  §  214 

.\Aa  =  Z.a\ 

.-.  A  APC  and  D^C  are  similar.  §  287 

( Two  rt.  A  are  similar  if  an  acute  Z  of  the  one  is  equal  to  an  acute  Z 
of  the  other.) 

.-.  CA  :  CD=CP:BC.  §  282 

.'.  CA  XBC=CDX  CP,  by  §  261.  q.e.d. 

This  theorem  may  be  omitted  without  destroying  the  sequence.  Props. 
XXIII  and  XXIV  are  occasionally  demanded  in  college  entrance  ex- 
aminations, but  they  are  not  necessary  for  proving  subsequent  propo- 
sitions or  for  any  of  the  exercises.  Teachers  may  therefore  use  their 
judgment  as  to  including  them. 


NUMERICAL  PROPERTIES  OF  FIGURES      181 
EXERCISE  47 

1.  The  tangents  to  two  intersecting  circles,  drawn  from  any 
point  in  their  common  chord  produced,  are  equal. 

2.  The  common  chord  of  two  intersecting  circles,  if  produced, 
bisects  their  common  tangents. 

3.  If  two  circles  are  tangent  externally,  the  common  internal 
tangent  bisects  the  two  common  external  tangents. 

4.  If  a  line  drawn  from  a  vertex  of  a  triangle  divides  the 
opposite  side  into  segments  proportional  to  the  adjacent  sides, 
the  line  bisects  the  angle  at  the  vertex. 

5.  If  three  circles  intersect  one  another,  the  common  chords 
are  concurrent. 

Let  two  of  the  chords,  AB  and  CD,  meet  at  0.  Join 
the  point  of  intersection  E  to  0,  and  suppose  that  EO 
produced  meets  the  same  two  circles  at  two  different 
points  P  and  Q.  Then  prove  that  OP  =  OQ  {^  299), 
and  hence  that  the  points  P  and  Q  coincide. 

6.  The  square  on  the  bisector  of  an  exterior  angle  of  a  triangle 
is  equal  to  the  product  of  the  segments  determined  by  this 
bisector  upon  the  opposite  side,  diminished  by  ^ 
the  product  of  the  other  two  sides.  /  '-/-^o^ 

Let  CD  bisect  the  exterior  ZBCH  of  the  A  ABC.     ^^ 
A  ADC  and  FBC  are  similar  (§  286).    Apply  §  303. 

7.  If  the  line  of  centers  of  two  circles  meets  the  circles  at  the 
consecutive  points  A,  B,  C,  D,  and  meets  the  common  external 
tangent  at  P,  then  PA  xPD  =  PBx  PC. 

8.  The  line  of  centers  of  two  circles  meets  the  common 
external  tangent  at  P,  and  a  secant  is  drawn  from  P,  cutting 
the  circles  at  the  consecutive  points  E,  F,  G,  II.  Prove  that 
PE  XPH=PFXPG. 

Draw  radii  to  the  points  of  contact,  and  to  E,  P,  G,  H.  Let  fall  Js  on 
PR  from  the  centers  of  the  ©.    The  various  pairs  of  A  are  similar. 


182  BOOK  III.    PLANE  GEOMETEY 

Proposition  XXV.    Problem 

306.  To  divide  a  given  line  into  parts  proportional 
to  any  number  of  given  lines. 

A         M'     n'  B 


^^ ^ V-^N 

\^^    \       \    \ 

\         "V     \    \ 

n-^\      \ 

"^ ^^\\ 

n p-\ 

p P-x 

Given  the  lines  AB^  m,  n,  and  p. 

Required  to  divide  AB  into  parts  proportional  to  m,  n,  and  p. 
Construction.    Draw  AX,  making  any  convenient  Z.  with  AB. 
On  AX  take  AM  equal  to  m, 
MN  equal  to  n,  and  NP  equal  to  j9. 
Draw  BP. 
From  N  draw  NN'  II  to  PB, 
and  from  M  draw  MM'  II  to  PB. 
Then    M'  and  iV'  are  the  division  points  required.         q.e.f. 
Proof.  Through  A  draw  a  line  II  to  PB. 

AM'  _M'N'      N'B 
AM  ~  MN  ~  NP' 

{Three  or  more  ||  lines  cut  off  proportional  intercepts  on  any  two 
transversals.) 

Substituting  7«,,  n,  and^^^  for  their  equals  AM,  MN,  and  NP, 

AM'      M'N'      N'B  ,       ^ 

we  have  = = Ax.  9 

7)1  n  p 

This  means  that  AB  has  been  divided  as  required.         q.e.d. 

In  like  manner,  we  may  divide  AB  into  parts  proportional  to  any 
number  of  given  lines. 


PROBLEMS  OF  CONSTRUCTION  183 

Proposition  XXVI.    Phoblem 

307.  To  find  the  fourth  proportional  to  three  given 
lines. 

A____yii_ B n_ G__^ 


\r 


Given  the  three  lines  m,  n,  and  p. 

Required  to  find  the  fourth  proportional  to  /«,  w,  and  p. 

Construction.    Draw  two  lines  ^A'  and  AY  containing  any 
convenient  angle. 

On  AX  take  AB  equal  to  m, 

and  take  BC  equal  to  n. 

On  ^  F  take  AD  equal  to  p. 

Draw  BD. 

From  C  draw  CE  II  to  BD,  meeting  A  Y  at  E. 

Then        DE  is  the  fourth  proportional  required.  q.e.f. 

Proof.  AB  :  BC  =  AD  :  DE.  §  273 

(If  a  line  is  drawn  through  tioo  sides  o/a  A  II  to  the  third  side,  it  divides 
the  two  sides  proportionally.) 

Substituting  7ti,  n,  and^  for  their  equals  AB,  BC,  and  AD, 
we  have  m:n=2^  '•  ^^-  ^^-  ^ 

Therefore  DE  is  the  fourth  proportional  to  m,  n,  and  p, 
by  §  259.  Q.E.D. 

308.  Corollary.    To  find  the    third  proportional  to  two 
given  lines. 

In  the  above  proof  take  m,  w,  n  as  the  given  lines  instead  of  m,  n,  p. 


184  BOOK  III.    PLANE  GEOMETEY 

Proposition  XXVII.    Problem: 

309.  To  find  the  mean  proportional  between  two  given 
lines. 


/ 

H 

\ 

\ 
1 

m 

! 

n 

! 

! 

A 

m 

n             B 

Given  the  two  lines  m  and  n. 

Required  to  find  the  mean  proportional  between  m  and  n. 

Construction.  Draw  any  line  AE,  and  on  AE  take  AC  equal 
to  m,  and  CB  equal  to  n. 

On  yli^  as  a  diameter  describe  a  semicircle. 

At  C  erect  the  _L  CH,  meeting  the  circle  at  //. 

Then  CH  is  the  mean  proportional  between  m  and  n.    q.e.f. 

Proof.  AC  :CH=CH:  CB.  §  297 

( The  A.  from  any  point  on  a  circle  to  a  diameter  is  the  mean 
proportional  between  the  segments  of  the  diameter.) 

Substituting  for  AC  and  CB  their  equals  m  and  n, 
we  have  m  :  CH=  CH :  n,  by  Ax.  9.  q.e.d. 

310.  Extreme  and  Mean  Ratio.  If  a  line  is  divided  into  two 
segments  such  that  one  segment  is  the  mean  proportional  be- 
tween the  whole  line  and  the  other  segment,  the  line  is  said  to 
be  divided  in  extreme  and  mean  ratio. 

E.g.  the  line  a  is  divided  in  extreme  and  mean  ratio,  if  a  segment  x 

is  found  such  that 

a  :  X  =  X  :  a  —  X. 

The  division  of  a  line  in  extreme  and  mean  ratio  is  often  called  the 
Golden  Section. 


PROBLEMS  OF  CONSTRUCTION      185 

Proposition  XXVIII.    Problem 
311.  To  divide  a  given  line  in  extreme  and  mean  ratio. 


:>Q 


^-1 


,.""nv 


A  C  B 

Given  the  line  AB. 

Recfiired  to  divide  AB  in  extreme  and  mean  ratio. 
Construction.    At  B  erect  a  ±  BE  equal  to  half  of  AB. 
From  £^  as  a  center,  with  a  radius  equal  to  EB,  describe  a  O. 

Draw  AE,  meeting  the  circle  at  F  and  G. 
On  ^^  take  ^C  equal  to  .IF. 

On  BA  produced  take  A  C '  equal  to  A  G. 

Then  AB  \^  divided  internally  at  C  and  externally  at  C  in 
extreme  and  mean  ratio. 

That  is,  AB  :  AC  =  AC  :  CB,  and  AB  :  AC  =  AC :  C'B.   q.e.f. 
Proof.  AG:AB  =  A B  :  A F.  §  302 


From  AG:AB  =  AB:AF, 
AG-AB:AB  = 

AB-AF:AF. 
.'.AG-FG:AB  = 

AB-AC:AC. 
.'.  AC:AB  =  CB:AC. 
.•.AB:AC  =  AC:CB, 
by  inversion,  §  266.  q.e.d. 


.'.  AB:AG  =  AF:  AB.     §  266 
.'.  AB-\-AG:  AG  = 

AF-\-AB:  AB. 

.\ABA-AC:AC  = 

AF  +  FG'.AB. 

.'.  CB:AC  =  AC:  AB. 
.-.  AB:AC  =  AC:C'B, 
by  §§261,  264.  q.e.d. 


186  BOOK  III.    PLANE  GEOMETRY 

Proposition  XXIX.    Pkoblem 

312.  Upon  a  given  line  correspondincj  to  a  given  side 
of  a  given  polygon,  to  construct  a  polygon  similar  to 
the  given  polygon. 

Y 

I 

I 

z     /i  \ 

\  /  I      \      ,^ 

\  /      --"  / 

Given  the  line  A^B^  and  the  polygon  ABCDE. 

Required  to  construct  on  A'B\  corresponding  to  AB,  a 
polygon  similar  to  the  polygon  ABCDE. 

Construction.    From  A  draw  the  diagonals  AD  and  AC. 
From  A ' draw  A 'X,  A'Y,  and  A 'Z,  making  Ax',  y\  and  z' equal 
respectively  to  A  x,  y,  and  «.  §  232 

From  B'  draw  a  line,  making  Z  B'  equal  to  Z  B, 

and  meeting  A'X  at  C'. 
From  C"  draw  a  line,  making  Z  D'C'B'  equal  to  Z  DCB, 

and  meeting  A'Y  sX  D'. 
From  i)'  draw  a  line,  making  ZE'D'C  equal  to  ZEDC, 

and  meeting  ^'Z  at  E'. 
Then  A'B'C'D'E'  is  the  required  polygon.  qe.f. 

Proof.  The  A  ABC  and  ^'^'C,  the  AACD  and  yl'C'/)',  and 
the  A  ADE  and  ^'Z>'J5:',  are  similar.  §  286 

Therefore  the  two  polygons  are  similar,  by  §  293.      q.e.d. 


EXERCISES  187 

EXERCISE  48 

1.  If  a  and  h  are  two  given  lines,  construct  a  line  equal  to  ar, 
where  x  =  'Wab.    Consider  the  special  case  of  «  =  2,  ^  =  3. 

2.  If  m  and  n  are  two  given  lines,  construct  a  line  equal  to  x, 
where  x  =  V 2  mn. 

3.  Determine  both  by  geometric  construction  and  arith- 
metically the  third  proportional  to  the  lines  1^  in.  and  2  in. 

4.  Determine  both  by  geometric  construction  and  arith- 
metically the  third  proportional  to  the  lines  4  in.  and  3  in. 

5.  Determine  both  by  geometric  construction  and  arith^ 
metically  the  fourth  proportional  to  the  lines  li  in.,  2  in.', 
and  21  in. 

6.  Determine  both  by  geometric  construction  and  arith- 
metically the  mean  proportional  between  the  lines  1.2  in. 
and  2.7  in. 

7.  Eind  geometrically  the  square  root  of  5.  Measure  the 
liiie  and  thus  determine  the  approximate  arithmetical  value. 

8.  A  map  is  drawn  to  the  scale  of  1  in.  to  50  mi.  How  far 
apart  are  two  places  that  are  2^\  in.  apart  on  the  map  ? 

9.  Eind  by  geometric  construction  and  arithmetically  the 
third  proportional  to  the  two  lines  ly^^  in.  and  2|  in. 

10.  Divide  a  line  1  in.  long  in  extreme  and  mean  ratio. 
Measure  the  two  segments  and  determine  their  lengths  to 
the  nearest  sixteenth  of  an  inch. 

11.  Divide  a  line  5  in.  long  in  extreme  and  mean  ratio. 
Measure  the  two  segments  and  determine  their  lengths  to 
the  nearest  sixteenth  of  an  inch. 

12.  Divide  a  line  6  in.  long  in  extreme  and  mean  ratio. 
Measure  the  two  segments  and  determine  their  lengths  to 
the  nearest  sixteenth  of  an  inch. 

The  propositions  on  this  page  are  taken  from  recent  college  entrance 
examination  papers. 


188 


BOOK  III.    PLANE  GEOMETRY 


13.  Through  a  given  point  P  within  a  given  circle  to  draw 
a  chord  AB  ^o  that  the  ratio  AP :  BP  shall  equal 
a  given  ratio  w  :  n. 

Draw  OPC  so  that  OP-.PC  =  n:  m. 
Draw  CA  equal  to  the  fourth  proportional  to  n,  w, 
and  the  radius  of  the  circle. 

14.  To  draw  two  lines  making  an  angle  of  60°,  and  to  con- 
struct all  the  circles  of  ^  in.  radius  that  are  tangent  to  both  lines. 

15.  To  draw  through  a  given  point  P  in  the 
arc  subtended  by  a  chord  ABsl  chord  which  shall 
be  bisected  by  AB.  e^ 

On  radius  OP  take  CD  equal  to  CP.  Draw  DE II  to  BA. 

16.  To  construct  two  circles  of  radii  ^  in.  and  1  in.  respec- 
tively, which  shall  be  tangent  externally,  and  to  construct  a 
third  circle  of  radius  3  in.,  which  shall  be  tangent  to  each  of 
these  two  circles  and  inclose  both  of  them. 

17.  To  draw  through  a  given  external 
point  P  a  secant  PAB  to  a  given  circle  so 
that  the  ratio  PA  :  AB  shall  equal  the  given 
ratio  7)1  :  n. 

Draw  the  tangent  PC.    Make  PD  .  DC  =  m  .  n.  PA  :  PC  =  PC  :  PB. 

18.  To  draw  through  a  given  external  point 
P  a  secant  PAB  to  a  given  circle  so  that 
AB^  =  PA  X  PB. 

19.  An  equilateral  triangle  ABC  is  2  in. 
on  a  side.  To  construct  a  circle  which  shall  be 
tangent  to  A  B  at  the  point  A  and  shall  pass  through  the  point  C. 

20.  To  draw  through  one  of  the  points 
of  intersection  of  two  circles  a  secant  so 
that  the  two  chords  that  are  formed  shall 
be  in  the  given  ratio  m  :  n. 


P^~, 


EXERCISES  189 

21.  In  a  circle  of  3  in.  radius  chords  are  drawn  through  a 
point  1  in.  from  the  center.  What  is  the  product  of  the  seg- 
ments of  these  chords  ? 

22.  The  chord  AB  is  S  in.  long,  and  it  is  produced  through 
B  to  the  point  P  so  that  PB  is  equal  to  12  in.  Find  the  tangent 
from  P. 

23.  Two  lines  AB  and  CD  intersect  at  0.  How  would  you 
ascertain,  by  measuring  OA,  OB,  OC,  and  OD,  whether  or  not 
the  four  points  A,  B,  C,  and  D  lie  on  the  same  circle  ? 

24.  This  figure  represents  an  instrument  for  finding  the 
centers  of  circular  plates  or  sections  of  shafts.  OC  is  a  ruler 
that  bisects  the  angle  A  OB,  and  A  0 
and  OB  are  equal.  Show  that,  if  A 
and  B  rest  on  the  circle,  OC  passes 
through  the  center,  and  that  by 
drawing  two  lines  the  center  can  be 
found. 

25.  If  three  circles  are  tangent  externally  each  to  the  other 
two,  the  tangents  at  their  points  of  contact  pass  through  the 
center  of  the  circle  inscribed  in  the  triangle  formed  by  joining 
the  centers  of  the  three  given  circles. 

26.  In  the  isosceles  triangle  ABC,  C  is  a  right  angle,  and  ^C 
is  4  in.  With  A  as  center  and  a  radius  2  in.  a  circle  is  described. 
Required  to  describe  another  circle  tangent  to  the  first  and  also 
tangent  to  BC  at  the  point  B. 

27.  Find  the  center  of  a  circle  of  ^  in.  radius,  so  drawn  in 
a  semicircle  of  radius  2  in.  as  to  be  tangent  to  the  semi- 
circle itself  and  to  its  diameter 

28.  To  inscribe  in  a  given  circle  a  triangle  similar  to  a  given 
triangle.  , ,  -    r  .c 

29.  To  draw  two  straight  line-segments,  having  given  their 
sum  and  their  ratio.  ....  ..        , 


190  BOOK  III.    PLANE  GEOMETRY 

EXERCISE  49 
Review  Questions 

1.  What  is  meant  by  ratio  ?  by  proportion  ? 

2.  If  a'.b=^c:  d,  write  four  other  proportions  involving 
these  quantities. 

3.  If  a  :  b  =  c :  d,  is  it  true  in  general  that  a-{-l:b  -\-l 
z=c:d?    Is  it  ever  true  ? 

4.  When  is  a  line  divided  harmonically  ?  The  bisectors  of 
what  angles  of  a  triangle  divide  the  opposite  side  harmonically  ? 

5.  What  are  the  two  conditions  necessary  for  the  similarity 
of  two  polygons  ? 

6.  Are  two  mutually  equiangular  triangles  similar  ?  Are 
two  mutually  equiangular  polygons  always  similar  ? 

7.  Are  two  triangles  similar  if  their  corresponding  sides 
are  proportional  ?  Are  two  polygons  always  similar  if  their 
corresponding  sides  are  proportional  ? 

8.  If  two  triangles  have  their  sides  respectively  parallel, 
are  they  similar  ?    Is  this  true  of  polygons  in  general  ? 

9.  If  two  triangles  have  their  sides  respectively  perpendicu- 
lar, are  they  similar  ?    Is  this  true  of  polygons  m  general  ? 

10.  Complete  in  two  ways :  The  perimeters  of  two  similar 
polygons  have  the  same  ratio  as  any  two  corresponding  ••• . 

11.  If  in  a  right  triangle  a  perpendicular  is  drawn  from 
the  vertex  of  the  right  angle  to  the  hypotenuse,  state  three 
geometric  truths  that  follow. 

12.  If  two  secants  intersect  outside,  on,  or  within  a  circle, 
what  geometric  truth  follows  ? 

"13.  How  would  you  proceed  to  divide  a  straight  line  into 
seven  equal  parts  ? 

- 14.  How.  would  you  proceed  to  find  the  square  root  of  7  by 
measuring  the  length  of  a  line  ?  '  ;.     ■ 


BOOK  IV 

AREAS  OF  POLYGONS 

313.  Unit  of  Surface.  A  square  the  side  of  which  is  a  unit  of 
length  is  called  a  unit  of  surface. 

Thus  a  square  that  is  1  inch  long  is  1  square  inch,  and  a  square  that  is 
1  mile  long  is  1  square  mile.  If  we  are  measuring  the  dimensions  of  a 
room  in  feet,  we  measure  the  surface  of  the  floor  in  square  feet.  In 
the  same  way  we  may  measure  the  page  of  this  book  in  square  inches 
and  the  area  of  a  state  in  square  miles. 

314.  Area  of  a  Surface.  The  measure  of  a  surface,  expressed 
in  units  of  surface,  is  called  its  area. 

If  a  room  is  20  feet  long  and  15  feet  wide,  the  floor  contains  300  square 
feet.  Therefore  the  area  of  the  floor  is  300  square  feet.  Usually  the  two 
sides  of  a  rectangle  are  not  commensurable,  although  by  means  of  frac- 
tions we  may  measure  them  to  any  required  degree  of  approximation. 
The  incommensurable  cases  in  theorems  like  Prop.  I  of  this  Book  may 
be  omitted  without  interfering  with  the  sequence  of  the  course. 

315.  Equivalent  Figures.  Plane  figures  that  have  equal  areas 
are  said  to  be  equivalent. 

In  propositions  relating  to  areas  the  words  rectangle^  triangle^  etc.,  are 
often  used  for  area  of  rectangle^  area  of  triangle^  etc. 

Since  congruent  figures  may  be  made  to  coincide,  congruent  figures 
are  manifestly  equivalent. 

Because  their  areas  are  equal,  equivalent  figures  are  frequently  spoken 
of  as  equal  figures.  The  symbol  =  is  used  both  for  "  equivalent "  and  for 
"  congruent,"  the  sense  determining  which  meaning  is  to  be  assigned  to  it. 
Occasionally  these  symbols  are  used  :  = ,  = ,  or  =  for  congruent,  =  for 
equal,  and  =o  for  equivalent. 

Since  the  word  congruent  means  "  identically  equal,"  the  word  equal  is 
often  used  to  mean ''equivalent." 

191 


192  BOOK  IV.    PLANE  GEOMETRY 

Proposition  I.    Theorem 

316.  Two  rectangles  having  equal  altitudes  are  to  each 
other  as  their  bases. 


D 


x  B  A       X 

Given  the  rectangles  -AC  and  AF^  having  equal  altitudes  AD. 

To  prove  that  \I}AC:\I\AF  =  base  AB  :  base  AE. 

Case  1.  When  AB  and  AE  are  commensurable. 

Proof.  Suppose  AB  and  AE  have  a  common  measure,  as  AX. 
Suppose  AX  '\?>  contained  m  times  in  AB  and  n  times  in  AE. 

Then  AB:AE  =  m:n. 

(For  m  and  n  are  the  numerical  measures  of  AB  and  AE.) 

Apply  ^X  as  a  unit  of  measure  to  ^^  and  AE,  and  at  the 
several  points  of  division  erect  Js. 

These  Js  are  all  ±  to  the  upper  bases,  §  97 

and  these  Js  are  all  equal.  §  128 

Since  to  each  base  equal  to  AX  there  is  one  rectangle, 
.'.  □  .4C  is  divided  into  m  rectangles, 
and  □  .4F  is  divided  into  n  rectangles.  §  119 

These  rectangles  are  all  congruent.  §  133 

.-,  ]Z\AC'.L^AF  =  m:n. 
.'.  {3  AC  '.n^AF^^AB'.AE,  by  Ax.  8.  q.e.d. 

In  this  proposition  we  again  meet  the  incommensurable  case,  as  on 
pages  116  and  157.  This  case  is  considered  on  page  193  and  may  be 
omitted  without  destroying  the  sequence  of  the  propositions. 


AREAS  OF  POLYGONS 


193 


Case  2.    When  AB  and  AE  are  incommensurable. 


D 


D 


Proof.  Divide  A  E  into  any  number  of  equal  parts,  and  apply 
one  of  these  parts  to  AB  as  many  times  as  AB  will  contain  it. 

Since  AB  and  AE  are  incommensurable,  a  certain  number  of 
these  parts  will  extend  from  A  to  some  point  P,  leaving  a  re- 
mainder PB  less  than  one  of  them.    Draw  PQ  _L  to  AB. 

□  .4Q      AP 
\Z\AF~  AE' 


Then 


Case  1 


By  increasing  the  number  of  equal  parts  into  which  AE  is 
divided  we  can  diminish  the  length  of  each,  and  therefore  can 
make  PB  less  than  any  assigned  positive  value,  however  small. 

Hence  PB  approaches  zero  as  a  limit,  as  the  number  of  parts 
of  AE  is  indefinitely  increased,  and  at  the  same  time  the  cor- 
responding □  PC  approaches  zero  as  a  limit.  §  204 

Therefore  AP  approaches  yl^  as  a  limit,  and  D^Q  ap- 
proaches \Z\  AC  as  a  limit. 

.*.  the  variable  — —  approaches  — —  as  a  limit, 
ACt  A.  E 

and  the  variable  7=— —  approaches  7=— —  as  a  limit. 

A  P  CH  A  Q 

But  — —  is  always  equal  to  ^^\  „?  as  AP  varies  in  value  and 


\3AF 


approaches  AB  as  a  limit. 


n\AC      AB    ^     ,  ^_ 


Case  1 


Q.E.D. 


317.  Corollary.    Two  rectangles  having  equal  bases  are  to 
each  other  as  their  altitudes. 


194  BOOK  IV.    PLANE  GEOMETRY 

Proposition  II.    Theorem 

318.  Tivo  rectangles  are  to  each  other  as  the  j^roducts 
of  their  bases  hy  their  altitudes. 


R 


b  U  b 

Given  the  rectangles  i?  and  R\  having  for  the  numerical  measure 
of  their  bases  b  and  h\  and  of  their  altitudes  a  and  a!  respectively. 

To  prove  that  -—  =  -—  • 

R      a'b' 

Proof.    Construct  the  rectangle  S,  with  its  base  equal  to  that 
of  R,  and  its  altitude  equal  to  that  of  R\ 

R_a 

'  R'~b'' 


Then 


and 


§317 
§316 


Since  we  are  considering  areas,  we  may  treat  R,  S,  and  S'  as 
numbers  and  take  the  products  of  the  corresponding  members 
of  these  equations.  ,-,         ^  §  272 


We  therefore  have 


n_ 

R' 


ab 


by  Ax.  3. 


Q.E.D. 


319.  Products  of  Lines.    When  we  speak  of  the  product  of  a 
and  b  we  mean  the  product  of  their  numerical 
values.    It  is  possible,  however,  to  think  of  a 
line  as  the  product  of  two  lines,  by  changing 
the  definition  of  multiplication.    Thus  in  this 
fig\ire  in  which  two  parallels  are  cut  by  two  intersecting  trans- 
versals, we  have  l:a  =  b  :  x.    Therefore  x  —  ah.    In  the  same 
way  we  may  find  xc^  or  abc^  the  product  of  three  lines. 


AREAS  OF  POLYGONS 
Proposition  III.    Theorem 


195 


320.  The  area  of  a  rectangle  is  equal  to  the  product 
of  its  base  hy  its  altitude. 


Q 


Given  the  rectangle  /?,  having  for  the  numerical  measure  of  its 
base  and  altitude  b  and  a  respectively. 

To  prove  that  the  area  of  R  —  ah. 

Proof.    Let  U  be  the  unit  of  surface.  §  313 


Then 


ab 


1x1 


ah. 


§318 


R 


But  J-  =  the  number  of  units  of  surface  in  R,  i.e.  the  area 
oiR.  §314 

.".  the  area  of  R  =  ab,  by  Ax.  8.  q.e.d. 

321.  Practical  Measures.  When  the  base  and  altitude  both 
contain  the  linear  unit  an  integral  number  of  times,  this  propo- 
sition is  rendered  evident  by  dividing  the  rectangle  into  squares, 
each  equal  to  the  unit  of  surface. 


Thus,  if  the  base  contains  seven  hnear 
units  and  the  altitude  four,  the  rectangle 
may  be  divided  into  twenty-eight  squares, 
each  equal  to  the  unit  of  surface.  Practi- 
cally this  is  the  way  in  which  we  conceive 
the  measure  of  all  rectangles.  Even  if  the 
sides  are  incommensurable,  we  cannot  determine  this  by  any  measuring 
instrument.  If  they  seem  to  be  incommensurable  with  a  unit  of  a 
thousandth  of  an  inch,  they  might  not  seem  to  be  incommensurable  with 
a  unit  of  a  millionth  of  an  inch. 


196  BOOK  IV.    PLANE  GEOMETRY 

EXERCISE  50 

1.  A  square  and  a  rectangle  have  equal  perimeters,  144 
yd.,  and  the  length  of  the  rectangle  is  five  times  the  breadth. 
Compare  the  areas  of  the  square  and  rectangle. 

2.  On  a  certain  map  the  linear  scale  is  1  in.  to  10  mi. 
How  many  acres  are  represented  by  a  square  |  in.  on  a  side  ? 

3.  Find  the  ratio  of  a  lot  90  ft.  long  by  60  ft.  wide  to  a 
field  40  rd.  long  by  20  rd.  wide. 

4.  Find  the  area  of  a  gravel  walk  3  ft.  6  in.  wide,  which 
surrounds  a  rectangular  plot  of  grass  40  ft.  long  and  25  ft. 
wide.    Make  a  drawing  to  scale  before  beginning  to  compute. 

5.  Find  the  number  of  square  inches  in  this 
cross  section  of  an  L  beam,  the  thickness  being  J  in.     ' 

6.  What  is  the  perimeter  of  a  square  field  that 
contains  exactly  an  acre  ?  '>^-2%-^ 

7.  A  machine  for  planing  iron  plates  planes  a  space  ^  in. 
wide  and  18  ft.  long  in  1  min.  How  long  will  it  take  to  plane 
a  plate  22  ft.  6  in.  long  and  4  ft.  6  in.  wide,  allowing  51  min. 
for  adjusting  the  machine  ? 

8.  How  many  tiles,  each  8  in.  square,  will  it  take  to  cover 
a  floor  24  ft.  8  in.  long  by  16  ft.  wide  ? 

9.  A  rectangle  having  an  area  of  48  sq.  in.  is  three  times 
as  long  as  wide.    What  are  the  dimensions  ? 

10.  The  length  of  a  rectangle  is  four  times  the  width.  If 
the  perimeter  is  60  ft.,  what  is  the  area? 

11.  From  two  adjacent  sides  of  a  rectangular  field  60  rd. 
long  and  40  rd.  wide  a  road  is  cut  4  rd.  wide.  How  many 
acres  are  cut  off  for  the  road  ? 

12.  From  one  end  of  a  rectangular  sheet  of  iron  10  in.  long 
a  square  piece  is  cut  off  leaving  25  sq.  in.  in  the  rest  of  the 
sheet.    How  wide  is  the  sheet  ? 


AREAS  OF  POLYGONS 
Proposition  IV.    Theorem 


197 


322.  Tlie  area  of  a  parallelogram  is  equal  to  the 
product  of  its  base  by  its  altitude. 

YD  X     G 


A  b  B  A  b  B 

Given  the  parallelogram  ABCD^  with  base  h  and  altitude  a. 
To  prove  that  the  area  of  the  CJABCD  =  ah. 
Proof.    From  B  draw  BX  1.  to  CD  or  to  CD  produced,  and 
from  A  draw  ^  F  J_  to  CD  produced. 

Then  ABXY  is  a  rectangle,  with  base  b  and  altitude  a. 
Since  AY=  BX,  2indAD  =  BC,  §  125 

.-.the  rt.  A  ADY  and  BCX  are  congruent.  §  89 

From  ABC  Y  take  the  A  BCX;  the  \Z1ABXY  is  left. 
From  ABC Y  take  the  A  ADY;  the  EJABCD  is  left. 

.-.  n\ABXY=  CJABCD.  Ax.  2 

But  the  area  of  the  □  A  BX  Y  =  ah.  §  320 

.*.  the  area  of  the  CJABCD  =  ah,  by  Ax.  8.         q.e.d. 

323.  C'oKOLLARY  1.  P aralUlograms  having  equal  bases  and 
equal  altitudes  are  equivalent. 

324.  Corollary  2.  Parallelograms  having  equal  bases  are 

to  each  other  as  their  altitudes;  parallelograms  having  equal 

altitudes  are  to  each  other  as  their  bases  ;  any  two  jmrallelo- 

grams  are  to  each  other  as  the  products  of  their  bases  by 

their  altitudes. 

This  was  regarded  as  veiy  interesting  by  the  ancients,  since  an  ignorant 
person  might  think  it  impossible  that  the  areas  of  two  parallelograms 
could  remain  the  same  although  their  perimeters  differed  without  limit. 


198  BOOK  IV.    PLANE  GEOMETRY 

Proposition  V.    Theorem 

325.  The   area  of  a    triangle  is  equal    to  half  the 
product  of  its  base  by  its  altitude. 


A  b  B 

Given  the  triangle  ABC,  with  altitude  a  and  base  b. 
To  prove  that  the  area  of  the  A  ABC  =  ^ab. 

Proof.    With  AB  and  BC  SiS  adjacent  sides   construct  the 

parallelogram  A  BCD.  §  238 

Then  A  ABC  =  i  EJABCD.  §  126 

But  the  area  of  the  EJABCD  =  ah.  §  322 

.*.  the  area  of  the  A  ABC  =  \  ab,  by  Ax.  4.  q.e.d. 

326.  Corollary  1.  Triangles  having  equal  bases  and  equal 
altitudes  are  equivalent. 

327.  Corollary  2.   Triangles  having  equal  bases  are  to  each 

other  as  their  altitudes  ;  triangles  having  equal  altitudes  are  to 

each  other  as  their  bases ;  any  two  triangles  are  to  each  other 

as  the  products  of  their  bases  by  their  altitudes. 

Has  this  been  proved  for  rectangles  ?  What  is  the  relation  of  a  triangle 
to  a  rectangle  of  equal  base  and  equal  altitude  ?  What  must  then  be  the 
relations  of  triangles  to  one  another  ? 

328.  Corollary  3.  The  product  of  the  sides  of  a  right  tri- 
angle is  equal  to  the  product  of  the  hypotenuse  by  the  altitude 
from  the  vertex  of  the  right  angle. 

How  is  the  area  of  a  right  triangle  found  in  terms  of  the  sides  of  the 
right  angle  ?  in  terms  of  the  hypotenuse  and  altitude  ?  How  do  these 
results  compare  ? 


AREAS  OF  POLYGONS 


199 


Proposition  VI.    Theorem 

329.  The  area  of  a  trapezoid  is  equal  to  half  the 
product  of  the  sum  of  its  bases  by  its  altitude. 


A  b  B 

Given  the  trapezoid  ABCD^  with  bases  b  and  V  and  altitude  a. 
To  prove  that  the  area  of  ABCD  ='^a(h-\-  h'}. 
Proof.  Di-aw  the  diagonal  A  C. 

Then     the  area  of  the  AABC  =  ^  ah, 
and  the  area  of  the  A  A  CZ)  =  ^  ah'.  §  325 

.*.  the  area  of  ABCD  =  ^  a  (^  +  h'),  by  Ax.  1.    q.e.d. 

330.  Corollary.    The  area  of  a  trapezoid  is  equal  to  the 

product  of  the  line  joining  the  mid-points  of  its  nonparallel 

sides  hy  its  altitude. 

How  is  the  line  joining  the  mid-points  of  the  nonparallel  sides  related 
to  the  sum  of  the  bases  (§  137)  ? 

331.  Area  of  an  Irregular  Polygon.  The  area  of  an  irregular 
polygon  may  be  found  by  dividing  the  polygon  into  triangles, 
and  then  finding  the  area  of  each 
of  these  triangles  separately. 

A  common  method  used  in  land  sur- 
veying is  as  follows :  Draw  the  longest 
diagonal,  and  let  fall  perpendiculars 
upon  this  diagonal  from  the  other  ver- 
tices of  the  polygon.  The  sum  of  the 
right  triangles,  rectahglep,  and  trapezoids  is  equivalent  to  the  polygon. 


200  BOOK  IV.    PLANE  GEOMETEY 

EXERCISE  51 

Find  the  areas  of  the  parallelograms  whose  bases  and 
altitudes  are  respectively  as  follows : 

1.  2.25  in.,  1 J  in.       3.  2.7  ft.,  1.2  ft.         5.  2  ft.  3  in.,  7  in. 

2.  3.44  in.,  li  in.       4.  5.6  ft.,  2.3  ft.         6.  3  ft.  6  in.,  2  ft. 

Find  the  areas  of  the  triangles  whose  bases  and  altitudes  ai'e 
respectively  as  follows  : 

7.  1.4  in.,  11  in.         9.  6^  ft.,  3  ft.  11.  1  ft.  6  in.,  8  in. 

8.  2.5  in.,  0.8  in.       10.  5.4  ft.,  1.2  ft.       12.  3  ft.  8  in.,  3  ft. 

Find  the  areas  of  the  trapezoids  ivhose  bases  are  the  first 
two  of  the  following  numbers,  and  whose  altitudes  are  the 
third  numbers  : 

13.  2  ft.,  1  ft.,  6  in.  15.  3  ft.  7  in.,  2  ft.,  14  in. 

14.  2^  ft.,  11  ft.,  9  in.  16.  5  ft.  6  in.,  3  ft.,  2  ft. 

Find  the  altitudes  of  the  parallelograms  whose  areas  and 
bases  are  respectively  as  follows : 

17.  10  sq.  in.,  5  in.     19.  28  sq.  ft.,  7  ft.      21.  30  sq.  ft.,  12  ft. 

18.  6sq.  in.,  6in.       20.  27  sq.  ft.,  6  ft.      22.  80  sq.  in.,  16  in. 

Find  the  altitudes  of  the  triangles  whose  areas  and  bases 
are  respectively  as  follows : 

23.  49  sq.  in.,  14  in.  25.  50  sq.  ft.,  10  ft.    27.  110  sq.  yd.,  10  yd. 

24.  48  sq.  in.,  12  in.    26.  160  sq.  ft.,  20  ft.  28.  176  sq.  yd.,  32  yd. 

Find  the  altitudes  of  the  trapezoids  whose  areas  and  bases 
are  respectively  as  follows : 

29.  33  sq.  in.,  5  in.,  6  in.  31.  13  sq.  ft.,  9  ft.,  5  ft. 

30.  15  sq.  in.,  4  in.,  6  in.  32.  70  sq.  yd.,  9  yd.,  11  yd.  ^. , 


AREAS  OF  POLYGONS  201 

Proposition  YII.    Theorem 

332.  The  areas  of  two  triangles  that  have  an  angle 
of  the  one  equal  to  an  angle  of  the  other  are  to  each 
other  as  the  2^roducts  of  the  sides  including  the  equal 
angles. 


Given  the  triangles  ABC  and  ADE^  with  the  common  angle  A, 

^  ,   ,  A  ABC     ABxAC 

To  prove  that  -7— rrrr.  =  ~7T^ 7~r,- 

^  AADE     ADxAE 

Proof.  Draw  BE. 


Then 


A  ABC      AC 


A  ABE      AE 


A  ABE      AB  ^  ^^„ 

{Triangleff  having  equal  altitudes  are  to  each  other  as  their  bases.) 

Since  we  are  considering  numerical  measures  of  area  and 
length,  we  may  treat  all  of  the  terms  of  these  proportions  as, 
numbers. 

Taking  the  product  of  the  first  members  and  the  product  of 
the  second  members  of  these  equations,  we  have 

AABE  X  A  ABC      AB  x  AC 


A  ADE  X  A  ABE      AD  X  AE 
That  is,  by  canceling  AABE,  we  have  the  proportion 


Ax.  3 


AABC      ABxAC 

= O.E.D 

AADE      ADXAE 


202  BOOK  IV.    PLANE  GEOMETRY 

Proposition  VIII.    Theorem 

333.  The  areas  of  tiuo  similar  triangles  are  to  each 
other  as  the  squares  on  any  two  corresjjonding  sides. 


A  B        A' 

Given  the  similar  triangles  ABC  and  A'B'C\ 

^              ^,   ^              A  ABC       AB' 
To  prove  that  .fp,^>  =  ■ 2 • 

Proof.    Since  the  triangles  are  similar,  Given 

.\AA=AA'.  §282 

_,  A  ABC         ABxAC 

^^^^  aJ^B^'^A'B'xA'C'  ^^^^ 

( The  areas  of  two  triangles  that  have  an  angle  of  the  one  equal  to 

an  angle  of  the  other  are  to  each  other  as  the  products 

of  the  sides  including  the  equal  angles.) 

.  AABC   _  AB_       AC_ 

Ihatis,  AA'B'C'- A'B'^A'C' 

But  -^  =  ~  •  §282 

A'B'      A'C 

{Similar  polygons  have  their  corresponding  sides  proportional.) 
Substituting  — — ,  for  its  equal  7777,'  we  have 

A  J:>  A  Ly 

AABC  _  AB_       AB_ 
AA'B'C'- A'B' ^  A'B''  ^^ 

AABC        AB^ 


AREAS  OF  POLYGONS  203 

Proposition  IX.    Theorem 

334.  The  areas  of  two  similar  polygons  are  to  each 
other  as  the  squares  on  any  tioo  corresponding  sides. 


Given  the  similar  polygons  ABODE  and  A'B^C'D'E\  of  area  5 
and  s'  respectively. 


To  prove  that 


s:s'  =  AB'':A'B'\ 


Proof.  By  drawing  all  the  diagonals  from  any  correspond- 
ing vertices  .4  and  A',  the  two  similar  polygons  are  divided 
into  similar  triangles. 

AADE       ad'       AACB       AC''       A  ABC 


AA'D'E'     Ji^r^     AA'C'D'     jJc' 
AADE        AACD 


That  is, 


AA'D'E'      AA'C'D' 
AADE-{-AACD-\-AABC 


•'     AA'B'C 

_  A  ABC 
~  AA'B'C' 

A  ABC 


AA'D'E'  +  A  A'C'D'  +  A  A'B'C      AA'B'C 

:2       -TT^2 


§292 

ab' 

A'B'' 

•  §333 

Ax.  8 

ab' 

§269 

s:s'  =  AB   :A'B'  ,  by  Ax.  11. 


Q.E.D. 


335.  Corollary  1.   The  areas  of  tivo  similar  polygons  are 
to  each  other  as  the  squares  on  any  two  corresponding  lines. 

336.  Corollary  2.    Corresponding  sides  of  hvo  similar  poly- 
gons have  the  same  ratio  as  the  square  roots  of  the  areas. 


204  BOOK  IV.    PLANE  GEOMETKY 

Proposition  X.    Theorem 

337.  Tlie  square  on  the  hypotenuse  of  a  right  triangle 
is  equivalent  to  the  sum  of  the  squares  on  the  other  tivo 
sides. 


/ 

R 


X     S 


Given  the  right  triangle  ABC^  with  AS  the  square  on  the  hypote- 
nuse, and  BN^  CQ  the  squares  on  the  other  two  sides. 

To  prove  that  AS=BN-\-CQ. 

Proof.  Draw  CX  through  C  II  to  ^.S".  §  233 

Draw  CR  and  BQ. 
Since  A  c  and  x  are  rt.  A,  the  Z  PCB  is  a  straight  angle,  §  34 


and  the  line  PCB  is  a  straight  line. 
Similarly,  the  line  A  CN  is  a  straight  line. 
In  the  A  ARC  and  ABQ, 

AR  =  AB, 
AC  =  AQ, 
and  ARAC  =  ZBAQ. 

{Each  is  the  sum  of  a  rt.  Z  and  the  Z  BAC.) 

.'.  A  ARC  is  congruent  to  A  ABQ. 
Furthermore  the  □  AX  is  double  the  A  ARC. 

{They  have  the  same  base  AR,  and  the  same  altitude  RX.) 


§43 


^65 
Ax.  1 

§68 
§325 


KUMERICAL  PROPERTIES  OF  FIGURES      205 

Again  the  square  CQ  is  double  the  AABQ.         §  325 
{They  have  the  same  base  AQ,  and  the  same  altitude  AC.) 

.'.  the  \I2AX  is  equivalent  to  the  square  CQ.         Ax.  3 

In  like  manner,  by  drawing  CS  and  AM,  it  may  be  proved 
that  the  rectangle  BX  is  equivalent  to  the  square  BN. 

Since  square  AS  =  nBX-^[D  AX,  Ax.  11 

.-.  AS  =  BN-{-  CQ,  by  Ax.  9.  q.e.d. 

The  first  proof  of  this  theorem  is  usually  attributed  to  Pythagoras  (about 
525  B.C.),  although  the  truth  of  the  proposition  was  known  earlier.  It  is 
one  of  the  most  important  propositions  of  geometry.  Various  proofs  may 
be  given,  but  the  on6  here  used  is  the  most  common.  This  proof  is  attrib- 
uted to  Euclid  (about  300  b.c),  a  famous  Greek  geometer. 

338.  Corollary  1.  The  square  on  either  side  of  a  right 
triangle  is  equivalent  to  the  difference  of  the  square  on  the 
hypotemise  and  the  square  on  the  other  side. 

339.  Corollary  2.  The  diagonal  and  a  side  of  a  square 
are  incommensurable.  jy  ^ 

For  AC^  =  AB^  +  BC^  =  2  AB^. 

.:  AC  =  ABV2. 

Since  V2  may  be  carried  to  as  many  decimal  places  as 
we  please,  but  cannot  be  exactly  expressed  as  a  rational 
fraction,  it  has  no  common  measure  with  1.    That  is,  AC :AB  =  V2,  an 
incommensurable  number. 

340.  Projection.  If  from  the  extremities  of  a  line-segment 
perpendiculars  are  let  fall  upon  another  line,  the  segment  thus 
cut  off  is  called  the  projection  of  the  first  line  upon  the  second. 

Thus  C^iy  is  the  projection  of  CD  upon  AB,  or  V  is  the  projection 
of  I  upon  AB. 

In  general  it  is  convenient  to  designate 
by  the  small  letter  a  the  side  of  a  triangle 
opposite  ZA,  and  so  for  the  other  sides;  to 
designate  the  projection  of  a  by  a';  and  to 


designate  the  height  (altitude)  by  A.  A       c^  d' 


206  BOOK  IV.    PLANE  GEOMETRY 

EXERCISE  52 

Given  the  sides  of  a  right  triangle  as  follotvs^  find  the 
hypotenuse  to  two  decimal  places: 

1.  30  ft.,  40  ft.         3.  20  ft.,  30  ft.  5.  2  ft.  6  in.,  3  ft. 

2.  45  ft.,  60  ft.         4.  1.5  in.,  2.5  in.  6.  3  ft.  8  in.,  2  ft. 

Given  the  hypotenuse  and  one  side  of  a  right  triangle  as 
follows,  find  the  other  side  to  two  decimal  places  : 

7.  50  ft,  40  ft.         9.  10  ft.,  6  ft.  11.  3  ft.  4  in.,  2  ft. 

8.  35  ft.,  21  ft.       10.  1.2  in.,  0.8  in.       12.  6  ft.  2  in.,  5  ft. 

13.  A  ladder  38  ft.  6  in.  long  is  placed  against  a  wall,  with 
its  foot  23.1  ft.  from  the  base  of  the  wall.  How  high  does  it 
reach  on  the  wall  ? 

14.  Find  the  altitude  of  an  equilateral  triangle  with  side  s. 

15.  Find  the  side  of  an  equilateral  triangle  with  altitude  h. 

16.  The  area  of  an  equilateral  triangle  with  side  s  is  i  s^  V3. 

17.  Find  the  length  of  the  longest  chord  and  of  the  shortest 
chord  that  can  be  drawn  through  a  point  1  ft.  from  the  center 
of  a  circle  whose  radius  is  20  in. 

18.  The  radius  of  a  circle  is  5  in.  Through  a  point  3  in. 
from  the  center  a  diameter  is  drawn,  and  also  a  chord  perpen- 
dicular to  the  diameter.  Find  the  length  of  this  chord,  and 
the  distance  (to  two  decimal  places)  from  one  end  of  the  chord 
to  the  ends  of  the  diameter. 

19.  In  this  figure   the   angle   C  is  a  >/l\ 
right  angle.    From  the  relations  AC  =              y^      \    \ 
AB  X  AF  (§  294)  ^d  C^=  AB  x  BF,  ^^  X  I       \^ 
show  that  lc'+  c]b%  Zb^                                          ^ 

20.  If  the  diagonals  of  a  quadrilateral  intersect  at  right 
angles,  the  sum  of  the  squares  on  one  pair  of  opposite  sides 
is  equivalent  to  the  sum  of  the  squares  on  the  other  pair. 


NUMERICAL  PROPERTIES  OF  FIGURES      207 

Proposition  XI.  Theorem 

341.  In  any  triangle  the  square  on  the  side  opposite 

an  acute  angle  is  equivalent  to  the  sum  of  the  squares 

on  the  other  two  sides  diminished  hy  ttvice  the  product 

of  one  of  those  sides  hy  the  projection  of  the  other  upon 

that  side. 

c 


Fig.  1 

Given  the  triangle  ABC^  A  being  an  acute  angle,  and  a'  and  V 
being  the  projections  of  a  and  b  resi)ectively  upon  c. 

To  prove  that  d^  =  ¥-\-c^—2  h'c. 

Proof.    If  D,  the  foot  of  the  _L  from  C,  falls  upon  c  (Fig.  1), 
a'  =  c-b'. 

If  D  falls  upon  c  produced  (Fig.  2), 

a'  =  b'-c. 
In  either  case,  by  squaring,  we  have 

a'2  =  Z»'2_f_^2_2^'c.  Ax.  5 

Adding  A^  to  each  side  of  this  equation,  we  have 

h^-^a'^  =  h^-{-b'^  +  c^-2b'c.  Ax.  1 

But  h^  +  «'2  =  a^,  and  h^  +  h""  =  b\  §  337 

Putting  a^  and  b'^  for  their  equals  in  the  above  equation,  we 


have 


a^  =  b^^c^-2b'c,  by  Ax.  9. 


Q.E.D. 


208  BOOK  IV.    PLANE  GEOMETRY 

Proposition  XII.    Theorem 

342.  In  any  obtuse  triangle  the  square  on  the  side 
opposite  the  obtuse  angle  is  equivalent  to  the  sum  of 
the  squares  on  the  other  tivo  sides  increased  by  tivice 
the  product  of  one  of  those  sides  by  the  projection 
of  the  other  upon  that  side. 


Given  the  obtuse  triangle  ABC^  A  being  the  obtuse  angle,  and  a' 
and  b'  the  projections  of  a  and  h  respectively  upon  c. 

To  prove  that  a^  =  b^  +  e'-\-2  b'c. 

Proof.  a'  =  b'  -\-c. 

Squaring,  a''' =  b'""  +  e" -{- 2  b'c. 

Adding  h^  to  each  side  of  this  equation,  we  have 

7,2  -f  ,,'2  =h^^  z,'2  4_  ^.2  _^  2  b'c. 
But  h^  +  a'^  =  a^,  and  Ii^  +  b'^  =  b\ 


Ax.  11 
Ax.  5 

Ax.  1 
§337 


Putting  (i^^iYidi  V^  for  their  equals  in  the  above  equation,  we  have 


a^  =  b^^^-^2b'c,\^j  Ky..  9. 


Q.E.D. 


Discussion.  By  the  Principle  of  Continuity  the  last  three  theorems 
may  be  included  in  one  theorem  by  letting  the  A  A  change  from  an 
acute  angle  to  a  right  angle  and  then  to  an  obtuse  angle.  Let  the 
student  explain. 

The  last  three  theorems  enable  us  to  compute  the  altitudes  of  a  tri- 
angle if  the  three  sides  are  known ;  for  in  Prop.  XII  we  can  find  6',  and 
from  h  and  V  we  can  find  h. 


NUMERICAL  PROPERTIES  OF  FIGURES      209 

EXERCISE  53 

Find  the  lengths^  to  two  decimal  places,  of  the  diagonals  of 
the  squares  whose  sides  are : 

1.  7  in.      2.  10  in.      3.  9.2  in.      4.  1  ft.  6  in.     5.  2  ft.  3  in. 

Find  the  lengths,  to  two  decimal  places,  of  the  sides  of 
the  squares  whose  diagonals  are : 

6.  4  in.       7.  8  in.       8.  5  ft.       9.    VE  in.       10.  2  ft.  6  in. 

11.  The  minute  hand  and  hour  hand  of  a  clock  are  6  in.  and 
4^  in.  long  respectively.  How  far  apart  are  the  ends  of  the 
hands  at  9  o'clock  ? 

12.  A  rectangle  whose  base  is  9  and  diagonal  15  has  the 
same  area  as  a  square  whose  side  is  x.    Find  the  value  of  x. 

13.  A  ring  is  screwed  into  a  ceiling  in  a  room  10  ft.  high. 
Two  rings  are  screwed  into  the  floor  at  points  5  ft.  and  12  ft. 
from  a  point  directly  beneath  the  one  in  the  ceiling.  Wires 
are  stretched  from  the  ceiling  ring  to  each  floor  ring.  How 
long  are  the  wires  ?    (Answer  to  two  decimal  places.) 

14.  The  sum  of  the  squares  on  the  segments  of  two  perpen- 
dicular chords  is  equivalent  to  the  square  on  the  diameter  of 
the  circle. 

If  AB,  CD  are  the  chords,  draw  the  diameter  BE,  and  draw  AC, 
ED,  BD.    Prove  that  AC  =  ED. 

15.  The  difference  of  the  squares  on  two  sides  of  a  triangle 
is  equivalent  to  the  difference  of  the  squares  on  the  segments 
of  the  third  side,  made  by  the  perpendicular  on  the  third 
side  from  the  opposite  vertex. 

16.  In  an  isosceles  triangle  the  square  on  one  of  the  equal 
sides  is  equivalent  to  the  square  on  any  line  drawn  from  the 
vertex  to  the  base,  increased  by  the  product  of  the  segments 
of  the  base. 


210 


BOOK  IV.    PLANE  GEOMETRY 


Proposition  XIII.    Theorem 

343.  The  sum  of  the  squares  on  tioo  sides  of  a  tri- 
angle is  equivalent  to  tivice  the  square  o?i  half  the  third 
side,  increased  hy  tivice  the  square  on  the  median 
upon  that  side. 

The  difference  of  the  squares  on  tioo  sides  of  a  tri- 
angle is  equivalent  to  tivice  the  j^^oduct  of  the  third 
side  hy  the  projection  of  the  median  upon  that  side. 


Given  the  triangle  ABC^  the  median  m,  and  mJ  the  projection  of 
m  upon  the  side  a.    Also  let  c  be  greater  than  b. 

To  prove  that    1.  c' +  6' =  2  .RE' +  2  w%- 
2.   c'-l>'  =  2am'. 

■  Proof.    The  Z  AMB  is  obtuse,  and  the  Z  CMA  is  acute.    §  116 
Since  c>b,  M  lies  between  B  and  D.  §  84 


Then 
and 


c^  =  BM  '  +  m^  -\-2BM-m', 

IP'  =  MC^  -\-m^-2MC  '  m'. 


§342 
§341 


Adding  these  equals,  and  observing  that  BiM  =  MC,  we  have 
c'i_^p  =  2  BM^  +  2  m^.  Ax.  1 

Subtracting  the  second  from  the  first,  we  have 

c^  —  b^  =  2  am',  by  Ax.  2.  q.e.d. 

Discussion.    Consider  the  proposition  when  c  =  h. 

This  theorem  may  be  omitted  without  interfering  with  the  regular 
sequence.  It  enables  us  to  compute  the  medians  when  the  three  sides 
are  known. 


EXERCISES  211 

EXERCISE  54 
1.  To  compute  the  area  of  a  triangle  in  terms  of  its  sides. 

c 


A       C       B     D 

At  least  one  of  the  angles  ^  or  B  is  acute.    Suppose  A  is  acute. 
In  the  A  A  DC,  h^  =  b"-AD^.  Why  ? 

In  the  A  ABC,  a^  =  ¥  -{- c^  -  2c  x  AD.  Why  ? 

62  4-  c2  -  a2 


Therefore  AD 

Hence  h^  =  b- 


2c 

(62  4.  c2  _  a2)2         4  ^2^2  _  (52  4.  c2  _  a2)2 


4c2  4c2 

_  (2  6c  +  62  ^  (.2  -  a2)  (2  6c  -  62  _  c2  +  a2) 
"         ~  4c2 

_  {(6  +  c)2  -  a2}  {a2  -  (6  -  c)2} 
~  4c2 

_  (g  +  6  +  c)  (6  +  c  —  g)  (g  +  6  —  c)  (g  —  6  +  c) 
-  4c2"' 

Let  g+6  +  c  =  2  5,  where  5  stands  for  semiperimeter. 

Then  6  +  c  —  g  =  g  +  6  +  c  —  2g  =  2s  —  2g  =  2(s  —  g). 

Similarly  •   g  +  6  —  c  =  2  (s  —  c), 

and  g-6  + c  =  2(s-6). 

^„      2  s  X  2  (.s  -  g)  X  2  (8  -  6)  X  2  (s  -  c) 

Hence  h^  = ^^ ^^ ^^ • 

4c2 

By  simplifying,  and  extracting  the  square  root, 

2    / 

hz=  -Vs{s  —  a)  (s  —  6)  (s  —  c). 


Hence  the  area  =  lch=  Vs  [s  —  a)  {s  —  6)  {s  —  c). 
For  example,  if  the  sides  arc  3,  4,  and  6, 


area  =  V6(6  -  3)  (6  -  4)  (6  -  5)  =  V6  •  3  •  2  =  6. 


212  BOOK  IV.    PLANE  GEOMETRY 

If  Ex.  1  has  been  studied,  find  the  areas,  to  two  decimal 
places,  of  the  triangles  whose  sides  are  : 

2.  4,  5,  6.         4.  6,  8,  10.         6.  7,  8,  11.  8.  1.2,  3,  2.1. 

3.  5,  6,  7.         5.  6,  8,  9.  7.  9,  10,  11.       9.  11,  12,  13. 

10.  To  compute  the  radius  of  the  circle  circumscribed  about 
a  triangle  in  terms  of  the  sides  of  the  triangle.  (Solve  only  if 
§  305  and  Ex.  1  have  been  taken.) 

Let  CD  be  a  diameter. 

By  §  305,  what  do  we  know  about  the  products  CA  x  BC  and  CD  x  CP  ? 

What  does  this  tell  us  of  ab  and  2  r  •  CP,  r  be-  

ing  the  radius  ?  X^      ^^^^^^ 

From  Ex.  1,  what  does  CP  equal  in  terms  of      /  ^^-^^^^/  i  ^ 
the  sides  ?  -^[^   c — 7^ — h~^^ 

Is  it  therefore  possible  to  show  that  I  /      ^-^^j 

abc  „  \      /'"^^         J 


4  Vs  (s  —  a)  (8  —  6)  (s  —  c)  .dV_      ^^ 

If  Exs.  1  and  10  have  been  studied,  compute  the  radii,  to 
two  decimal  places,  of  the  circles  circumscribed  about  the 
triangles  whose  sides  are : 

11.  3,  4,  5.      12.  27,  36,  45.      13.  7,  9,  11.      14.  10,  11,  12. 

15.  To  compute  the  medians  of  a  triangle  in  terms  of  its  sides. 
Omit  if  §  343  has  not  been  taken.    What  do  we  know 
about  a^  +  6^  as  compared  with  2  m^  -j-  2  ( -  J  ? 
From  this  relation  show  that 


If  Ex.  15  has  been  studied,  compute  the  three  medians,  to 
two  decimal  places,  of  the  triangles  whose  sides  are  : 

16.  3,  4,  5.  17.  6,  8,  10.  18.  6,  7,  8.         19.  7,  9,  11. 

20.  If  the  sides  of  a  triangle  are  7,  9,  and  11,  is  the  angle 
opposite  the  side  11  right,  acute,  or  obtuse  ? 


EXERCISES 


213 


1^ 

I 

-I . 

H 


.H 


\C 


21.  The  square  constructed  upon  the  sum  of  two  lines  is 
equivalent  to  the  sum  of  the  squares  constructed  upon  these 
two  lines,  increased  by  twice  the  rectangle  of  these  lines. 

Given  the  two  lines  AB  and  BC,  and  AC  their  sum.  Construct  the 
squares  AKGC  and  ADEB  upon  AC  and  AB  respec- 
tively. Produce  BE  and  DE  to  meet  KG  and  CG  in  H  ^  B  c 
and  F  respectively.  Then  we  have  the  square  EHGF, 
with  sides  each  equal  to  BC.  Hence  the  square  AKGC 
is  the  sum  of  the  squares  ADEB  and  EHGF,  and  the 
rectangles  DKHE  and  BEFC. 

This  proves  geometrically  the  algebraic  formula 
(a  +  6)2  =  a2  +  2  a6  +  b^. 

22.  The  square  constructed  upon  the  difference  of  two  lines 
is  equivalent  to  the  sum  of  the  squares  constructed  upon  these 
two  lines,  diminished  by  twice  their  rectangle. 

Given  the  two  lines  AB  and  AC,  and  BC  their  dif- 
ference. Construct  the  square  AGFB  upon  AB,  the 
square  A  CKH  upon  A  C,  and  the  square  CDEB  upon 
BC.  Produce  ED  to  meet  AG  in  L.  The  dimensions 
of  the  rectangles  LGFE  and  HLDK  are  AB  and  AC, 
and  the  square  CDEB  is  the  difference  between  the 
whole  figure  and  the  sum  of  these  rectangles. 

This  proves  geometrically  the  algebraic  formula 
(a  _  6)2  =  a2  -  2  ab  +  b^. 

23.  The  difference  between  the  squares  constructed  upon' 
two  lines  is  equivalent  to  the  rectangle  of  the  sum  and  differ- 
ence of  these  lines.  /         k 

Given  the  squares  ^BD J?  and  CBFG,  constructed 
upon  AB  and  BC.  The  difference  between  these 
squares  is  the  polygon  ACGFDE,  which  is  composed 
of  the  rectangles  ACHE  and  GFDH.  Produce  AE 
and  CH  to  /  and  K  respectively,  making  EI  and  HK 
each  equal  to  BC,  and  draw  IK.  The  difference  be-  ^ 
tween  the  squares  ABDE  and  CBFG  is  then  equiva- 
lent to  the  rectangle  ACKI,  with  dimensions  AB  +  BC,  and  AB 

This  proves  geometrically  the  algebraic  formula 
a2- 62  =  (a +  6)  (a -6). 


. 1> 


\ 

H 

G 

BC. 


214  BOOK  IV.    PLANE  GEOMETRY 

Proposition  XIV.    Problem 

344.  To  construct  a  square  equivalent  to  the  sum  of 
tivo  given  squai^es. 


Given  the  two  squares,  i?  and  i?'. 

Required  to  construct  a  square  equivalent  to  E  -\-  E'. 
Construction.    Construct  the  rt.  Z.4.  §  228 

On  the  sides  of  Z  .4,  take  AB,  or  c,  equal  to  a  side  of  R',  and 
AC,  or  b,  equal  to  a  side  of  R,  and  draw  BC,  or  a. 

Construct  the  square  S,  having  a  side  equal  to  BC. 

Then  ^  is  the  square  required.  q.e.f. 

Proof.  a-=Z;'--|-6'2.  §337 

{The  square  on  the  hypotenuse  of  a  rt.  A  is  equivalent  to  the  sum  of 
the  squares  on  the  other  two  sides.) 

.'.  S  =  R-{-R',  by  Ax.  9.  q.e.d. 

345.  Corollary  1.   To  construct  a  square  equivalent  to  the 

difference  of  two  given  squares. 

We  may  easily  reverse  the  above  construction  by  first  dravi^ing  .c. 
then  erecting  a  ±  at  ^,  and  then  with  a  radius  a  fixing  the  point  C. 

346.  Corollary  2.   To  construct  a  square  equivalent  to  the 

Slim  of  three  given  squares. 

If  a  side  of  the  third  square  is  d,  we  may  erect  a  perpendicular  from 
C  to  the  line  J5C,  take  CD  equal  to  d,  and  join  D  and  B. 

Discussion.  It  is  evident  that  we  can  continue  this  process  indefi- 
nitely, and  thus  construct  a  square  equivalent  to  the  sum  of  any  number 
of  given  squares. 


PKOBLEMS  OF  CONSTKUCTION 


215 


Proposition  XV.    Problem 

347.  To  construct  ci  j^ohjgon  similar   to   two  given 
similar  polygons  and  equivalent  to  their  sicrn. 


R" 


Given  the  two  similar  polygons  R  and  i?'. 

Required  to  construct  a  polygon  similar  to  R  and  R\  and 
equivalent  to  R -\-  R\ 

Construction.    Construct  the  rt  AO.  §  228 

Let  s  and  s'  be  corresponding  sides  of  R  and  R\ 
On  the  sides  of  /.O,  take  OX  equal  to  s',  and  OY  equal  to  s. 

Draw  XY,  and  take  s"  equal  to  XY. 
Upon  s",  corresponding  to  s,  construct  W  similar  to  11.      §  312 
Then  it"  is  the  polygon  required.  q.e.f. 

Proof.  OF^ -\-'0X'^  =  XY^.  §  337 

Putting  for  OY,  OX,  and  XF their  equals  s,  s',  and  s",  we  have 


But 
and 

By  addition, 


R"~s"^'' 

R"~s"^' 
R-^R' 


s'-\- 


R' 


1. 


:.  R"  =  R  +  PJ,  by  Ax.  3. 


Ax.  9 

§334 
Ax.  1 

Q.E.D. 


216 


BOOK  IV.    PLANE  GEOMETRY 


Pkoposition  XVI.    Problem 

348.  To  construct  a  triangle  equivalent  to  a  given 
polygon. 


Given  the  polygon  ABCDEF. 

Required  to  construct  a  triangle  equivalent  to  ABCDEF. 

Construction.    Let   B,    C,  and  D  be  any  three   consecutive 
vertices  of  the  polygon.    Draw  the  diagonal  DB. 

From  V  draw  a  line  II  to  DB.  §  233 

Produce  AB  to  meet  this  line  at  Q,  and  draw  DQ. 

Again,  draw  EQ,  and  from  D  draw  a  line  II  to  EQ,  meeting 
AB  produced  at  R,  and  draw  ER. 

In  like  manner  continue  to  reduce  the  number  of  sides  of 
the  polygon  until  we  obtain  the  A  EPR. 

Then  A  £JP/?  is  the  triangle  required.  q.e.f. 

Proof.    The  polygon  AQDEF  has   one   side    less   than    the 
polygon  ABCDEF. 

Furthermore,  in  the  two  polygons,  the  ipSii'tABDEF  is  common, 
and  the  A  BQD  =  A  BCD.  §  326 

(For  the  base  DB  is  common,  and  their  vertices  C  and  Q  are  in 
the  line  CQ  II  to  the  base.) 

.-.  AQDEF  =  ABCDEF.  Ax.  1 

In  like  manner  it  may  be  proved  that 

AREF=  AQDEF,  and  EPR  =  AREF.  Q.e.d. 


PROBLEMS  OF  CONSTRUCTION 


217 


Proposition  XVII.    Problem 

349.  To   construct   a   square   equivalent  to  a   given 

parallelogram. 

p 


N 


M 


A         h  B 

Given  the  parallelogram  ABCD. 

Required  to  construct  a  square  equivalent  to  the  EJABCD. 

Construction.    Upon  any  convenient  line  take  NO  equal  to  a, 
and  OM  equal  to  b,  the  altitude  and  base  respectively  of  O 

ABCD. 

Upon  NM  as  a  diameter  describe  a  semicircle. 

At  O  erect  OP  _L  to  NM,  meeting  the  circle  at  P.    §  228 

Construct  the  square  S,  having  a  side  equal  to  OP. 

Then  S  is  the  square  required.  q.e.f. 


Proof. 

That  is, 


But 

and 


NO  :OP  =  OP:  OM. 
.'.  0P^=N0X0M. 
OF''=ah. 
S  =  0P\ 


§297 
§261 
Ax.  9 


EJABCD  =  ab.  §322 

.-.  S=CJABCD,  by  Ax.  9.  q.e.d. 

350.  Corollary  1.   To  construct  a  square  equivalent  to  a 

given  triangle. 

Take  for  a  side  of  the  square  the  mean  proportional  between  the  base 
and  half  the  altitude  of  the  triangle. 

351.  Corollary  2.   To  construct  a  square  equivalent  to  a 

given  polygon. 

First  reduce  the  polygon  to  an  equivalent  triangle,  and  then  construct 
a  square  equivalent  to  the  triangle. 


218  BOOK  IV.    PLANE  GEOMETRY 

Proposition  XVIII.    Problem 

352.  To  construct  a  parallelogram  equivalent  to  a 
given  square,  and  having  the  sum  of  its  base  and 
altitude  equal  to  a  given  line. 


A      Q  B 

Given  the  square  S,  and  the  line  AB. 

Required  to  construct  a  O  equivalent  to  S^  with  the  sum  of 
its  base  and  altitude  equal  to  AB. 

Construction.    Upon  AB  as  a  diameter  describe  a  semicircle. 

At  A  erect  AC  A-  to  AB  and  equal  to  a  side  of  the  given 

square  S.  §  228 

Draw  CD  II  to  AB,  cutting  the  circle  at  P.  §  233 

Draw  PQ  _L  to  AB.  §  227 

Then  any  CJ,  as  P,  having  AQ  for  its  altitude  and  QB  for 


its  base  is  equiva 

lent  to  .S. 

Q.E.F. 

Proof. 

AQ:PQ  =  PQ:  QB. 

§297 

.'.Pq''=AQx  QB. 

§261 

Furthermore 

PQ  is  II  toC^. 

§95 

.\PQ  =  CA. 

§127 

.'.PQ^  =  CA\ 

Ax.  5 

.•..4QX  QB=CA\ 

Ax.  8 

But 

P  =  AQx  QB, 

§322 

and 

S=zCA\ 

§320 

.'.  P=S,  by  Ax.  8. 

Q.E.D. 

Thus  is  solved  geometrically  the  algebraic  problem,  given  x  +  y  =  a, 
xy  =  6,  to  find  x  and  y. 


PKOBLEMS  OF  CONSTRUCTION 


219 


Pkoposition  XIX.    Problem 

353.  To  construct  a  parallelogram  equivalent    to  a 
given  square,  and  having  the  difference  of  its  base  and 
altitude  equal  to  a  given  line, 
c 


.E^ 


-\B 


Given  the  square  S,  and  the  line  AB. 

Required  to  construct  a  O  equivalent  to  S,  ivith  the  differ- 
ence of  its  base  and  altitude  equal  to  AB. 

Construction.    Upon  .45  as  a  diameter  describe  a  circle. 

From  A  draw  A  C\  tangent  to  the  circle,  and  equal  to  a  side 
of  the  given  square  S. 

Through  the  center  of  the  circle  draw  CD  intersecting  the 
circle  at  E  and  D. 

Then  any  O,  as  P,  having  CD  for  its  base  and  CE  for  its 
altitude,  is  equivalent  to  aS.  q.  e.  f. 

Proof.                     CD :  CA  =  CA  :  CE.  §  302 

.'.CA^  =  CDxCE,  §261 
and  the  difference  between  CD  and  CE  is  the  diameter  of  the 
circle,  that  is,  AB. 

But                                   P=CDx  CE,  §322 

and                                       S=CA''.  §320 

.'.  P=  S,hy  Ax.  8.  Q. E.  D. 

Thus  is  solved  geometrically  the  algebraic  problem,  given  x  —  y  =  a, 
xy  =  &,  to  find  x  and  y. 


220  BOOK  IV.    PLANE  GEOMETRY 

Proposition  XX.    Problem 

354.  To  construct  a  j)olygon  similar  to  a  given  poly- 
gon and  equivalent  to  another  given  polygon. 


Given  the  polygons  P  and  Q. 

Required  to  construct  a  polygon  similar  to  P  and  equiva- 
lent to  Q. 

Construction.    Construct  squares  equivalent  to  P  and  Q,  §  351 
and  let  m  and  n  respectively  denote  their  sides. 

Let  s  be  any  side  of  P. 

Find  s\  the  fourth  proportional  to  m^  n,  and  s.  §  307 
Upon  s\  corresponding  to  s, 

construct  a  polygon  P'  similar  to  the  polygon  P.  §  312 

Then                  P'  is  the  polygon  required.  q.e.f. 

Proof.    Since                   m :  n  =  s :  s',  Const. 

.•.m^:7i'  =  s^:s'\  §270 

But                           P  =  711^,  and  Q  =  n^.  Const. 

.'.P:Q  =  s'':s'\  Ax.  9 

But                                 P'.P^  =  s' :  s'l  §  334 

{The  areas  of  two  similar  polygons  are  to  each  other  as  the  squares  on 
any  two  corresponding  sides.) 

,'.  P:Q  =  P:P'.  Ax.  8 

.'.P'=Q.  §263 

.*.  P',  being  similar  to  P,  is  the  polygon  required,    q.e.d. 


PROBLEMS  OF  CONSTRUCTION 


221 


Proposition  XXI.    Problem 

355.  To  construct  a  square  which  shall  have  a  given 
ratio  to  a  given  square. 


s 

D 
#-^.    / 

Given  the  square  S,  and  the  ratio  — . 

m 

Required  to  construct  a  square  which  shall  he  to  S  as  n  is 
to  m. 

Construction.    Take  AB  equal  to  a  side  of  S,  and  draw  /IF, 
making  any  convenient  angle  with  AB. 

On  A  Y  take  AE  equal  to  m  units  and  EF  equal  to  n  units. 

Draw  EB. 
From  F  draw  a  line  II  to  EB,  meeting  AB  produced  at  C.  §  233 

On  .4  C  as  a  diameter  describe  a  semicircle. 
At  B  erect  BD  J_  to  ^  C,  meeting  the  semicircle  at  D. 
Then  BD  is  a  side  of  the  square  required 

Proof.    Denote  AB  hj  a,  BC  hj  b,  and  BD  by  x. 


§228 

Q.E.F. 


Then 


But 


By  inversion, 


a  :  x^x : 

b. 

.  a:b  =  a^ 

:x\ 

a\h  —  m 

:  n. 

a^  :x^  =  m 

:  n. 

x^  :a^  =  n: 

Tfl. 

§297 
§271 
§273 
Ax.  8 
§266 


Hence  the  square  on  BD  will  have  the  same  ratio  to  S  as 
n  has  to  m.  q.e.d. 


222  BOOK  IV.    PLANE  GEOMETRY 

Proposition  XXII.    Problem 

356.  To  construct  a  jjolycjon  similar  to  a  given  j)oly- 
gon  and  having  a  given  ratio  to  it. 


Given  the  polygon  P  and  the  ratio  —  • 

772 

Required  to  construct  a  polygon  similar  to  P,  which  shall  he 
to  P  as  n  is  to  m. 

Construction.    Let  s  be  any  side  of  P. 

Draw  a  line  ,s-',  such  that  the  square  on  *•'  shall  be  to  the 
square  on  ^  as  t^  is  to  m.  §  355 

Upon  *•'  as  a  side  corresponding  to  s  construct  the  polygon 
P'  similar  to  P.  §  312 

{Upon  a  given  line  corresponding  to  a  given  side  of  a  given  polygon^ 
to  construct  a  polygon  similar  to  the  given  polygon.) 

Then  P'  is  the  polygon  required.  q.  e.  f. 

Proof.  P':P  =  s"':s\  §334 

( The  areas  of  two  similar  polygons  are  to  each  other  as  the  squares 
on  any  two  corresponding  sides.) 

But  s'^ :  s'^  —  n:  m.  Const. 

Therefore  P^ :  P  =  n  :  m,  hj  Ax.  8.  q.e.d. 

This  problem  enables  us  to  construct  a  square  that  is  twice  a  given 
square  or  half  a  given  square,  to  construct  an  equilateral  triangle  that 
shall  be  any  number  of  times  a  given  equilateral  triangle,  and  in  general 
to  enlarge  or  to  reduce  any  figure  in  a  given  ratio.  An  architect's  drawl- 
ing, for  example,  might  need  to  be  enlarged  so  as  to  be  double  the  area 
of  the  original,  and  the  scale  could  be  found  by  this  method. 


EXERCISES  223 

EXERCISE  55 
Problems  of  Computation 

1.  The  sides  of  a  triangle  are  0.7  in.,  0.6  in.,  and  0.7  in. 
respectively.    Is  the  largest  angle  acute,  right,  or  obtuse  ? 

2.  The  sides  of  a  triangle  are  5.1  in.,  6.8  in.,  and  8.5  in. 
respectively.    Is  the  largest  angle  acute,  right,  or  obtuse  ? 

3.  Find  the  area  of  an  isosceles  triangle  whose  perimeter 
is  14  in.  and  base  4  in.    (One  decimal  place.) 

4.  Find  the  area  of  an  equilateral  triangle  whose  perimeter 
is  18  in.    (One  decimal  place.) 

5.  Find  the  area  of  a  right  triangle,  the  hypotenuse  being 
1.7  in.  and  one  of  the  other  sides  being  0.8  in. 

6.  Find  the  ratio  of  the  altitudes  of  two  triangles  of  equal 
area,  the  base  of  one  being  1.5  in.  and  that  of  the  other  4.5  in. 

7.  The  bases  of  a  trapezoid  are  34  in.  and  30  in.,  and  the 
altitude  is  2  in.    Find  the  side  of  a  square  having  the  same  area. 

8.  What  is  the  area  of  the  isosceles  right  triangle  in  which 
the  hypotenuse  is  V2  ? 

9.  AVhat  is  the  area  of  the  isosceles  right  triangle  in  which 
the  hypotenuse  is  7  V2  ? 

10.  If  the  side  of  an  equilateral  triangle  is  2  V3,  what  is  the 
altitude  of  the  triangle  ?  the  area  of  the  triangle  ? 

11.  If  the  side  of  an  equilateral  triangle  is  1  ft.,  what  is  the 
area  of  the  triangle  ? 

12.  If  the  area  of  an  equilateral  triangle  is  43.3  sq.  in.,  what 
is  the  base  of  the  triangle  ?    (Take  V3  =  1.732.) 

13.  The  sides  of  a  triangle  are  2.8  in.,  3.5  in.,  and  2.1  in. 
respectively.  Draw  the  figure  carefully  and  see  what  kind  of 
a  triangle  it  is.  Verify  this  conclusion  by  applying  a  geometric 
test,  and  find  the  area  of  the  triangle. 


224  BOOK  lY.    PLANE  GEOMETRY 

EXERCISE  56 
Theorems 

1.  The  area  of  a  rhombus  is  equal  to  half  the  product  of 
its  diagonals. 

2.  Two  triangles  are  equivalent  if  the  base  of  the  first 
is  equal  to  half  the  altitude  of  the  second,  and  the  altitude 
of  the  first  is  equal  to  twice  the  base  of  the  second. 

3.  The  area  of  a  circumscribed  polygon  is  equal  to  half  the 
product  of  its  perimeter  by  the  radius  of  the  inscribed  circle. 

4.  Two  parallelograms  are  equivalent  if  their  altitudes  are 
reciprocally  proportional  to  their  bases. 

5.  If  equilateral  triangles  are  constructed  on  the  sides  of  a 
right  triangle,  the  triangle  on  the  hypotenuse  is  equivalent  to 
the  sum  of  the  triangles  on  the  other  two  sides. 

6.  If  similar  polygons  are  constructed  on  the  sides  of  a 
right  triangle,  as  corresponding  sides,  the  polygon  on  the 
hypotenuse  is  equivalent  to  the  sum  of  the  polygons  on  the 
other  two  sides. 

Ex.  6  is  one  of  the  general  forms  of  the  Pythagorean  Theorem. 

7.  If  lines  are  drawn  from  any  point  within  a  parallelogram 
to  the  four  vertices,  the  sum  of  either  pair  of  triangles  with 
parallel  bases  is  equivalent  to  the  sum  of  the  other  pair. 

8.  Every  line  drawn  through  the  intersection  of  the  diag- 
onals of  a  parallelogram  bisects  the  parallelogram. 

9.  The  line  that  bisects  the  bases  of  a  trapezoid  divides  the 
trapezoid  into  two  equivalent  parts. 

10.  If  either  diagonal  of  a  trapezoid  bisects  it,  the  trapezoid 
is  a  parallelogram. 

11.  The  triangle  formed  by  two  lines  drawn  from  the  mid- 
point of  either  of  the  nonparallel  sides  of  a  trapezoid  to  the 
opposite  vertices  is  equivalent  to  half  the  trapezoid. 


EXERCISES  225 

EXERCISE  57 
Problems  of  Construction 

1.  Given  a  square,  to  construct  a  square  of  half  its  area. 

2.  To  construct  a  right  triangle  equivalent  to  a  given 
oblique  triangle. 

3.  To  construct  a  triangle  equivalent  to  the  sum  of  two 
given  triangles. 

4.  To  construct  a  triangle  equivalent  to  a  given  triangle, 
and  having  one  side  equal  to  a  given  line. 

5.  To  construct  a  rectangle  equivalent  to  a  given  parallelo- 
gram, and  having  its  altitude  equal  to  a  given  line. 

6.  To  construct  a  right  triangle  equivalent  to  a  given  tri- 
angle, and  having  one  of  the  sides  of  the  right  angle  equal  to 
a  given  line, 

7.  To  construct  a  right  triangle  equivalent  to  a  given  tri- 
angle, and  having  its  hypotenuse  equal  to  a  given  line. 

8.  To  divide  a  given  triangle  into  two  equivalent  parts  by 
a  line  through  a  given  point  P  in  the  base. 

9.  To  draw  from  a  given  point  P  in  the  base  ylJ5  of  a  tri- 
angle ABC  2^  line  to  AC  produced,  so  that  it  may  be  bisected 
by  BC. 

10.  To  find  a  point  within  a  given  triangle  such  that  the 
lines  from  this  point  to  the  vertices  shall  divide  the  triangle 
into  three  equivalent  triangles. 

11.  To  divide  a  given  triangle  into  two  equivalent  parts  by 
a  line  parallel  to  one  of  the  sides. 

12.  Through  a  given  point  to  draw  a  line  so  that  the  seg- 
ments intercepted  between  the  point  and  perpendiculars  drawn 
to  the  line  from  two  other  given  points  may  have  a  given  ratio. 

13.  To  find  a  point  such  that  the  perpendiculars  from  it  to 
the  sides  of  a  given  triangle  shall  be  in  the  ratio  p,  q,  r. 


226  BOOK  IV.    PLANE  GEOMETRY 

EXERCISE  58 
Review  Questions 

1.  What  is  meant  by  the  area  of  a  surface  ?    Illustrate. 

2.  What  is  the  difference  between  equivalent  figures  and 
congruent  figures  ? 

3.  State  two  propositions  relating  to  the  ratio  of  one 
rectangle  to  another. 

4.  Given  the  base  and  altitude  of  a  rectangle,  how  is  the  area 
found  ?    Given  the  area  and  base,  how  is  the  altitude  found  ? 

5.  How  do  you  justify  the  expression,  "  the  product  of  two 
lines  "  ?  "  the  quotient  of  an  area  by  a  line  "  ? 

6.  Can  a  triangle  with  a  perimeter  of  10  in.  have  the  same 
area  as  one  with  a  perimeter  of  1  in.  ?  Is  the  same  answer 
true  for  two  squares  ? 

7.  Can  a  parallelogram  with  a  perimeter  of  10  in.  have  the 
same  area  as  a  rectangle  with  a  perimeter  of  1  in.  ?  Is  the 
same  answer  true  for  two  rectangles  ? 

8.  Explain  how  the  area  of  an  irregular  field  with  straight 
sides  may  be  found  by  the  use  of  the  theorems  of  Book  IV. 

9.  A  triangle  has  two  sides  5  and  6,  including  an  angle  of 
70°,  and  another  triangle  has  two  sides  2  and  7^,  including  an 
angle  of  70°.    What  is  the  ratio  of  the  areas  of  the  triangles  ? 

-  10.  Two  similar  triangles  have  two  corresponding  sides  5  in. 
and  15  in.  respectively.  The  larger  triangle  has  how  many 
times  the  area  of  the  smaller  ? 

11.  Given  the  hypotenuse  of  an  isosceles  right  triangle,  how 
do  you  proceed  to  find  the  area  ? 

12.  Given  three  sides  of  a  triangle,  what  test  can  you  apply 
to  determine  whether  or  not  it  is  a  right  triangle  ? 

13.  Suppose  you  wish  to  construct  a  square  equivalent  to  a 
given  polygon,  how  do  you  proceed  ? 


BOOK  V 

REGULAR  POLYGONS  AND  CIRCLES 

357.  Regular  Polygon.  A  polygon  that  is  both  equiangular 
and  equilateral  is  called  a  regular  polygon. 

Familiar  examples  of  regular  polygons  are  the  equilateral  triangle  and 
the  square. 

It  is  proved  in  Prop.  I  (§  362)  that  a  circle  may  be  circumscribed  about, 
and  a  circle  may  be  inscribed  in,  any  regular  polygon,  and  that  these 
circles  are  concentric  (§  188) . 

358.  Radius.  The  radius  of  the  circle  cir- 
cumscribed about  a  regular  polygon  is  called 
the  radius  of  the  polygon. 

In  this  figure  r  is  the  radius  of  the  polygon. 

359.  Apothem.  The  radius  of  the  circle  inscribed  in  a  regular 
polygon  is  called  the  apothem  of  the  polygon. 

In  the  figure  a  is  the  apothem  of  the  polygon.  The  apothem  is  evi- 
dently perpendicular  to  the  side  of  the  regular  polygon  (§  185) . 

360.  Center.  The  common  center  of  the  circles  circumscribed 
about  and  inscribed  in  a  regular  polygon  is  called  the  center  of 
the  polygon. 

In  the  figure  O  is  the  center  of  the  polygon. 

361.  Angle  at  the  Center.  The  angle  between  the  radii  drawn 
to  the  extremities  of  any  side  of  a  regular  polygon  is  called  the 
angle  at  the  center  of  the  polygon. 

In  the  figure  m  is  the  angle  at  the  center  of  the  polygon.  It  is  evidently 
subtended  by  the  chord  which  is  the  side  of  the  inscribed  polygon. 

227 


228 


BOOK  V.    PLANE  GEOMETRY 


Proposition  I.    Theorem 

362.  A  circle  may  he  circumscribed  about,  and  a  circle 
may  he  inscribed  in,  any  regular  polygon. 


Given  the  regular  polygon  ABCDE. 

To  prove  that  1.  a  circle  may  he  circumscribed  about  ABCDE ; 
2.  a  circle  may  be  inscribed  in  ABCDE. 

Proof.  1.  Let  O  be  the  center  of  the  circle  which  may  be 
passed  through  the  three  vertices  A,  B,  and  C.  §  190 

Draw  OA,  OB,  OC,  OD. 

Then                                   OB  =  OC,  §162 

and                                          AB  =  CD.  §357 

Furthermore               Z  CBA  =  Z  DCB,  i  357 

and                                    Z  CBO  =  Z  OCB.  §  74 

.\Z0BA  =  ZDC0.  Ax.  2 

.-.A  OAB  is  congruent  to  AOCD.  §  68 

.'.OA=OD.  §67 

Therefore  the  circle  that  passes  through  A,  B,  C,  passes  also 
through  D. 

In  like  manner  it  may  be  proved  that  the  circle  that  passes 
through  B,  C,  and  D  passes  also  through  E;  and  so  on. 

Therefore  the  circle  described  with  0  as  a  center  and  OA  as  a 
radius  will  be  circumscribed  about  the  polygon,  by  §  205.    q.  e.  d. 


REGULAR  POLYGOKS  AND  CIRCLES         229 

Proof.    2.  Let  O  be  the  center  of  the  circumscribed  circle. 

D 


Since  the  sides  of  the  regular  polygon  are  equal  chords  of 
the  circumscribed  circle,  they  are  equally  distant  from  the 
center.  §  178 

Therefore  the  circle  described  with  0  as  a  center,  with  the 
perpendicular  from  O  to  a  side  of  the  polygon  as  a  radius,  will 
be  inscribed  in  the  polygon,  by  §  205.  q.e.d. 

363.  Corollary  1.  The  radius  drawn  to  any  vertex  of  a 
regular  polygon  bisects  the  angle  at  the  vertex. 

364.  Corollary  2.   The  angles  at  the  center  of  any  regular 

polygon  are  equal,  and  each  is  supplementary  to  an  interior 

angle  of  the  polygon. 

For  the  angles  at  the  center  are  corresponding  angles  of  congruent 
triangles.  If  M  is  the  mid-point  of  AB^  then  since  the  A  MOB  and  OBM 
are  complementary  what  can  we  say  of  their  doubles,  AOB  and  CBA  ? 

365.  Corollary  3.  An  equilateral  polygon  inscribed  in  a 
circle  is  a  regular  polygon. 

Why  are  the  angles  also  equal  ? 

366.  Corollary  4.  An  equiangular  polygon  circumscribed 
about  a  circle  is  a  regular  polygon. 

By  joining  consecutive  points  of  contact  of  the  sides  of  the  polygon 
can  you  show  that  certain  isosceles  triangles  are  congruent,  and  thus 
prove  the  polygon  equilateral  ? 


230  BOOK  V.    PLANE  GEOMETRY 

Proposition  II.    Theorem 

367.  If  a  circle  is  divided  into  any  number  of  equal 
arcs,  the  chords  joining  the  successive  points  of  division 
form  a  regular  inscribed  polygon;  and  the  tangents 
drawn  at  the  points  of  division  form  a  regidar  circum- 
scribed polygon. 


Given  a  circle  divided  into  equal  arcs  by  ^,  5,  C,  Z),  and  E^  AB^ 
BCj  CD,  DE,  and  EA  being  chords,  and  PQ,  QR,  RS,  ST,  and  TP 
being  tangents  at  B,  C,  D,  E,  and  A  respectively. 

To  prove  that  1.  ABODE  is  a  regular  polygon  ; 
2.  PQRST  is  a  regular  polygon. 

Proof.    1.  Since  the  arcs  are  equal  by  construction, 

.-.  AB  =  BC  =  CD  =  DE  =  EA.  §  170 

.'.  ABODE  is  a  regular  polygon.  §  365 

{An  equilateral  polygon  inscribed  in  a  circle  is  a  regular  polygon.) 

Proof.    2.         AP  =  Z.Q  =  ZR  =  ZS  =  ZT.  §221 

{An  Z  formed  by  two  tangents  is  measured  by  half  the  difference  of 
the  intercepted  arcs.) 

.*.  PQRST  is  a  regular  polygon.  §  366 

{An  equiangular  polygon  circumscribed  about  a  circle  is  a 

regular  polygon.)  Q.  E.  D. 


KEGULAR  POLYGONS  AND  CIRCLES 


231 


368.  Corollary  1.  Taiigents  to  a  circle  at  the  vertices  of 
a  regular  inscribed  polygon  form  a  regular  circumscribed  poly- 
gon of  the  same  number  of  sides. 

369.  Corollary  2.  Tangents  to  a  circle  at  the  mid-points 
of  the  arcs  subtended  by  the  sides  of  a  regular  inscribed 
polygon  form  a  regular  circumscribed 
polygon^  whose  sides  are  parallel  to  the 
sides  of  the  inscribed  polygon  and  whose 
vertices  lie  on  the  radii  (^produced^  of  ^^ 
the  inscribed  polygon. 

For  two  corresponding  sides,  AB  and  A'B'^ 
are  perpendicular  to  OM  (§§  176,  185),  and  are 
parallel  (§  95) ;  and  the  tangents  MB'  and  iVJB', 

intersecting  at  a  point  equidistant  from  OM  and  ON  (§  192),  intersect  upon 
the  bisector  of  the  Z  MON  {%  152)  ;  that  is,  upon  the  radius  OB  (§  363). 

370.  Corollary  3.  Lines  draum  from  each 
vertex  of  a  regular  polygon  to  the  7nid-points 
of  the  adjacent  arcs  subtended  by  the  sides  of 
the  polygon  form  a  regular  inscribed  polygon 
of  double  the  number  of  sides. 

371.  Corollary  4.    Tangents  at   the   mid- 
points of  the  arcs  between  adjacent  points  of 
contact  of  the  sides  of  a  regular  circumscribed 
polygon  form  a  regular  circumscribed  polygon  ^^ 
of  double  the  number  of  sides.  j 

372.  Corollary  5.  The  perimeter  of  a  regular  inscribed 
polygon  is  less  than  that  of  a  regular  inscribed  polygon  of 
double  the  number  of  sides;  and  the  perimeter  of  a  regular 
circumscribed  polygori  is  greater  than  that  of  a  regular  cir- 
cumscribed polygofi  of  double  the  number  of  sides. 


232  BOOK  V.    PLANE  GEOMETRY 

EXERCISE  59 

1.  Find  the  radius  of  the  square  whose  side  is  5  in. 

2.  Find  the  side  of  the  square  whose  radius  is  7  in. 

3.  Find  the  radius  of  the  equilateral  triangle  whose  side 
is  2  in.         ^ 

4.  Find  the  side  of  the  equilateral  triangle  whose  radius 
is  3  in. 

5.  Find  the  apothem  of  the  equilateral  triangle  whose  side 
is  VS  in. 

6.  Find  the  side  of  the  equilateral  triangle  whose  apothem 
is  2  V3  in. 

7.  How  many  degrees  are  there  in  the  angle  at  the  center 
of  an  equiangular  triangle  ?  of  a  regular  hexagon  ? 

8.  Given  an  equilateral  triangle  inscribed  in  a  circle,  to 
circumscribe  an  equilateral  triangle  about  the  circle. 

9.  Given  an  equilateral  triangle  inscribed  in  a  circle,  to  in- 
scribe a  regular  hexagon  in  the  circle,  and  to  circumscribe  a 
regular  hexagon  about  the  circle. 

10.  Given  a  square  inscribed  in  a  circle,  to  inscribe  a  regular 
octagon  in  the  circle,  and  to  circumscribe  a  regular  octagon 
about  the  circle. 

11.  How  many  degrees  are  there  in  the  angle  at  the  center 
of  a  regular  octagon  ?  in  each  angle  of  a  regular  octagon  ?  in 
the  sum  of  these  two  angles  ? 

12.  What  is  the  area  of  the  square  inscribed  in  a  circle  of 
radius  2  in.? 

13.  The  apothem  of  an  equilateral  triangle  is  equal  to  half 
the  radius. 

14.  Prove  that  the  apothem  of  an  equilateral  triangle  is  equal 
to  one  fourth  the  diameter  of  the  circumscribed  circle.  From 
this  show  how  an  equilateral  triangle  may  be  inscribed  in  a  circle. 


REGULAE  POLYGONS  AND  CIRCLES 
Proposition  III.    Theorem 


233 


373.   Tivo  regular  polygons  of  the  same  numher  of 
sides  are  similar. 


§145 


Given  the  regular  polygons  P  and  P',  each  having  n  sides. 

To  prove  that        P  and  P'  are  similar. 

2(71-2) 
Proof.  Each  angle  of  either  polygon  = rt.  A. 

2  (n  —  2) 
{Each  Z  of  a  regular  polygon  of  n  sides  is  equal  to  — rt.  A.) 

Hence  the  polygons  P  and  P'  are  mutually  equiangular. 

Furthermore,     '.'  AB  =  BC  =  CD  =  DE  =  EA, 

and  A'B'  =  B'C  =  CD'  =  D'E'  =  E'A',  §  357 

AB  _  BC  _  CD  _  DE  _  EA 
'  '  A'B'~  B'C'~  CD' ~  D'E' ~  E'A'' 

Hence  the  polygons  have  their  corresponding  sides  propor- 
tional and  their  corresponding  angles  equal. 

Therefore  the  two  polygons  are  similar,  by  §  282.  q.e.d. 

374.  Corollary.    The  areas  of  two  regular  polygons  of  the 

same  numher  of  sides  are  to  each  other  as  the  squares  on  any 

two  corresponding  sides. 

Por  the  areas  of  two  similar  polygons  are  to  each  other  as  tlie  squares 
on  any  two  corresponding  sides  (§  334),  and  two  regular  polygons  of  the 
same  number  of  sides  are  similar. 


Ax.  4 


234 


BOOK  V.    PLANE  GEOMETEY 
Proposition  IV.    Theorem 


375.  The  perimeters  of  two  regular  p)olygons  of  the 
same  number  of  sides  are  to  each  other  as  their  radii, 
and  also  as  their  apothems. 


A      MB  A'      M'      B' 

Given  the  regular  polygons  with  perimeters  p  and  p\  radii  r 
and  r',  apothems  a  and  a',  and  centers  O  and  0'  respectively. 

To  prove  that  p  :  p'  =  r  :  r'  =  a:  a'. 

Proof.    Since  the  polygons  are  similar, 

.\p:2)'  =  AB:A'B'. 

Furthermore  in  the  isosceles  AOAB  and  O'A'B', 

and  OA  :  OB  =  O'A' :  O'B'. 

{For  each  of  these  ratios  equals  1.) 

.-.the  AOAB  and  O'A'B'  are  similar. 
.'.AB:A'B'  =  r:r'. 
Also  A  A  MO  and  A'M'O'  are  similar. 

.'.  r  :r'  =  a  :  a'. 
.-.  j9  i]^'  =  r  :  r'  =  a  :  a',  by  Ax.  8. 

376.  Corollary.  The  areas  of  two  regular  polygons  of  the 
same  number  of  sides  are  to  each  other  as  the  squares  on  the 
radii  of  the  circumscribed  circles,  and  also  as  the  squares  on 
the  radii  of  the  inscribed  circles. 


§373 
§291 

§364 
Ax.  8 

§288 
§282 
§286 

§282 

Q.  E.  D. 


KEGULAR  POLYGONS  AND  CIRCLES  235 

EXERCISE  60 

1.  Eind  the  ratio  of  the  perimeters  and  the  ratio  of  the 
areas  of  two  regular  hexagons,  their  sides  being  2  in.  and 
4  in.  respectively. 

2.  Find  the  ratio  of  the  perimeters  and  the  ratio  of  the 
areas  of  two  regular  octagons,  their  sides  having  the  ratio  2  :  6. 

3.  Eind  the  ratio  of  the  perimeters  of  two  squares  whose 
areas  are  121  sq.  in.  and  30i  sq.  in.  respectively. 

4.  Eind  the  ratio  of  the  perimeters  and  the  ratio  of  the 
areas  of  two  equilateral  triangles  whose  altitudes  are  3  in. 
and  12  in.  respectively. 

5.  The  area  of  one  equiangular  triangle  is  nine  times  that 
of  another.    Required  the  ratio  of  their  altitudes. 

6.  The  area  of  the  cross  section  of  a  steel  beam  1  in.  thick 
is  12  sq.  in.  What  is  the  area  of  the  cross  section  of  a  beam  of 
the  same  proportions  and  1^  in.  thick  ? 

7.  Squares  are  inscribed  in  two  circles  of  radii  2  in.  and 
6  in.  respectively.  Eind  the  ratio  of  the  areas  of  the  squares, 
and  also  the  ratio  of  the  perimeters. 

8.  Squares  are  inscribed  in  two  circles  of  radii  2  in.  and 
8  in.  respectively,  and  on  the  sides  of  these  squares  equi- 
lateral triangles  are  constructed.  What  is  the  ratio  of  the 
areas  of  these  triangles  ? 

9.  A  round  log  a  foot  in  diameter  is  sawed  so  as  to  have 
the  cross  section  the  largest  square  possible.  What  is  the  area 
of  this  square  ?  What  would  be  the  area  of  the  cross  section 
of  the  square  beam  cut  from  a  log  of  half  this  diameter  ? 

10.  Every  equiangular  polygon  inscribed  in  a  circle  is  regular 
if  it  has  an  odd  number  of  sides. 

11.  Every  equilateral  polygon  circumscribed  about  a  circle 
is  regular  if  it  has  an  odd  number  of  sides. 


236  BOOK  V.    PLANE  GEOMETRY 

Proposition  V.    Theorem 

377.  If  the  number  of  sides  of  a  regular  inscribed 
polygon  is  indefinitely  increased,  the  ai^otheni  of  the 
polygon  approaches  the  radius  of  the  circle  as  its  limit. 


Given  a  regular  polygon  of  n  sides  inscribed  in  the  circle  of 
radius  OA^  s  being  one  side  and  a  the  apothem. 

To  prove  that  a  approaches  r  as  a  limits  if  n  is  increased 
indefinitely. 

Proof.    We  know  that         a<r.  §  86 

Then  since                      r~a<AM,  §  112 

and                                         AM<s,  §174 

.\r  —  a<,s.  Ax.  10 

If  n  is  taken  sufficiently  great,  s,  and  consequently  r—  a,  can 
be  made  less  than  any  assigned  positive  value,  however  small. 

Since  r—a  can  become  and  remain  less  than  any  assigned 
positive  value  by  increasing  n,  it  follows  that 

r  is  the  limit  of  a,  by  §  204.  q.'e.d. 

378.  Corollary.  //  the  number  of  sides  of  a  regular  in- 
scribed polygon  is  indefinitely  increased^  the  square  on  the 
apothem  approaches  the  square  on  the  radius  of  the  circle 
as  its  limit. 

For  r2  -  a2  ^  AM'^.  §  338 

Therefore  by  taking  n  sufficiently  great,  s,  and  consequently  AM,  and 
consequently  r^  —  a^,  approaches  zero  as  its  limit. 


REGULAR  POLYGONS  AND  CIRCLES         237 

Proposition  VI.    Theorem 

379.  An  arc  of  a  circle  is  less  than  a  line  of  any  kind 
that  envelops  it  on  the  convex  side  and  has  the  same 
extremities. 


Given  BCA  an  arc  of  a  circle,  AB  being  its  chord. 

To  prove  that  the  arc  BCA  is  less  than  a  line  of  any  hind 
that  envelops  this  arc  and  terminates  at  A  and  B. 

Proof.  Of  all  the  lines  that  can  be  drawn,  each  to  include 
the  area  ABC  between  itself  and  the  chord  AB,  there  must  be 
at  least  one  shortest  line ;  for  all  the  lines  are  not  equal. 

Let  BDA  be  any  kind  of  line  enveloping  BCA  as  stated. 

The  enveloping  line  BDA  cannot  be  the  shortest ;  for  draw- 
ing ECF  tangent  to  the  arc  BCA  at  any  point  C,  the  line 
BFCEA<BFDEA,  since  FCE<  FDE.  Post.  3 

In  like  manner  it  can  be  shown  that  no  other  enveloping 
line  can  be  the  shortest. 

.*.  BCA  is  shorter  than  any  enveloping  line.        q.e.d. 

380.  Corollary.  A  circle  is  less  than  the  perimeter  of  any 
polygon  circumscribed  about  it. 

381.  Circle  as  a  Limit.   From  Prop.  VI  we  may  now  assume : 

1.  The  circle  is  the  limit  which  the  perimeters  of  regular  in- 
scribed polygons  and  of  similar  circumscribed  polygons  approach, 
if  the  number  of  sides  of  the  polygons  is  indefinitely  increased. 

2.  The  area  of  the  circle  is  the  limit  which  the  areas  of  the 
inscribed  and  circumscribed  polygons  approach. 


238 


BOOK  V.    PLANE  GEOMETEY 


Proposition  YII.    Theorem 

382.   Tico  circumferences  hcwe  the  same  ratio  as  their 
radii. 


Given  the  circles  with  circumferences  c  and  c',  and  radii  r  and 
r'  respectively. 

To  prove  that  c:  c'  =  r  :  r'. 

Proof.  Inscribe  in  the  circles  two  similar  regular  polygons, 
and  denote  their  perimeters  by  ^  and^>'. 

Then  p:2)'  =  r:7''.  §375 

Conceive  the  number  of  sides  of  these  regular  polygons  to 
be  indefinitely  increased,  the  polygons  continuing  similar. 

Thenp  and  2^'  will  have  c  and  c'  as  limits.  §  381 

But  the  ratio  p  :p'  will  always  be  equal  to  the  ratio  ?• :  r' 
.'.  jyr'  =2)'t'. 
.'.  cr'  =  c'r. 
,'.c:c'  =  r:r',  by  §264. 

383.  Corollary  1.   The  ratio  of  any  circle  to  its  diameter 
is  constant. 

Why  does  c  :  c'=  2  r  :  2  r'  ?   Then  why  does  c  :  2  r  =  c' :  2  r'  ? 

384.  The  Symbol  tt.    The  constant  ratio  of  a  circle  to  its 
diameter  is  represented  by  the  Greek  letter  rrr  (pi). 

385.  Corollary  2.    In  any  circle  c=2irr. 


§375 
§261 
§207 

Q.E.D. 


For,  by  definition,  17  = 

2r 


REGULAR  POLYGONS  AND  CIRCLES  239 

Proposition  VIII.    Theorem 

386.   The  area  of  a  regular  polygon  is  equal  to  half 
the  product  of  its  apothem  hy  its  perimeter. 


A       M      B 


Given  the  regular  polygon  ABCDEF,  with  apothem  a,  perimeter />, 
and  area  s. 

To  prove  that  *  —  2  ^P' 

Proof.  Draw  the  radii  OA,  OB,  OC,  etc.,  to  the  successive 
vertices  of  the  polygon. 

The  polygon  is  then  divided  into  as  many  triangles  as  it 
has  sides. 

The  apothem  is  the  common  altitude  of  these  A,  and  the 
area  of  each  A  is  equal  to  \a  multiplied  by  the  base.        §  325 

Hence  the  area  of  all  the  triangles  is  equal  to  \  a  multiplied 
by  the  sum  of  all  the  bases.  Ax.  1 

But  the  sum  of  the  areas  of  all  the  triangles  is  equal  to  the 
area  of  the  polygon.  Ax.  11 

And  the  sum  of  all  the  bases  of  the  triangles  is  equal  to  the 
perimeter  of  the  polygon.  Ax.  11 

.-.  s  =  ^  ap.  Q.E.  D. 

387.  Similar  Parts.  In  different  circles  similar  arcs,  similar 
sectors,  and  similar  segments  are  such  arcs,  sectors,  and  seg- 
ments as  correspond  to  equal  angles  at  the  center. 

For  example,  two  arcs  of  30°  in  different  circles  are  similar  arcs,  and 
the  corresponding  sectors  are  similar  sectors. 


240  BOOK  V.    PLANE  GEOMETRY 

Proposition  IX.    Theorem 

388.   The  area  of  a  circle  is  equal  to  half  the  product 
of  its  radius  hy  its  circumference. 


Given  a  circle  with  radius  r,  circumference  c,  and  area  s. 
To  prove  that 


s  =  1  re. 

Proof.  Circumscribe  any   regular   polygon   of  n  sides,  and 

denote  the  perimete*-  of  this  polygon  by  p  and  its  area  by  s\ 

Then  since  r  is  its  apothem,  s'  =  ^  77^.  §  386 
Conceive  n  to  be  indefinitely  increased. 

Then  since  p  approaches  c  as  its  limit,  §  381 
and  r  is  constant, 
-'■  \rp  approaches  \rc  as  its  limit. 

Also                    s'  approaches  s  as  its  limit.  §  381 

But                               s'  =  \rp  always.  §  386 

.\s  =  lrc,  by  §  207.  q.e.d. 

389.  Corollary  1.   The  area  of  a  circle  is  equal  to  ir  times 
the  square  on  its  radius. 

For  the  area  of  the  O  =  \rc  =  \r  x  ^irr  —  tit^. 

390.  Corollary  2.  The  areas  of  two  circles  are  to  each  other 
as  the  squares  on  their  radii. 

391.  Corollary  3.   The  area  of  a  sector  is  equal  to  half  the 
product  of  its  radius  hy  its  are. 

area  of  sector      arc  of  sector 


For 


area  of  circle  circle 


EEGULAR  POLYGONS  AND  CIRCLES         241 

EXERCISE  61 

1.  Two  circles  are  constructed  with  radii  1^^  in.  and  4^  in. 
respectively.  The  circumference  of  the  second  is  how  many 
times  that  of  the  first? 

2.  The  circumference  of  one  circle  is  three  times  that  of 
another.  The  square  on  the  radius  of  the  first  is  how  many 
times  the  square  on  the  radius  of  the  second? 

3.  The  circumference  of  one  circle  is  2^  times  that  of 
another.  The  equilateral  triangle  constructed  on  the  diameter 
of  the  first  has  how  many  times  the  area  of  the  equilateral 
triangle  constructed  on  the  diameter  of  the  second  ? 

4.  A  circle  with  a  diameter  of  5  in.  has  a  circumference  of 
15.708  in.  What  is  the  circumference  of  a  pipe  that  has  a 
diameter  of  2  in.  ? 

5.  A  wheel  with  a  circumference  of  4  ft.  has  a  diameter 
of  1.27  ft.,  expressed  to  two  decimal  places.  What  is  the  cir- 
cumference of  a  wheel  with  a  diameter  of  1.58|  ft.  ? 

6.  A  regular  hexagon  is  2  in.  on  a  side.  Find  its  apothem 
and  its  area  to  two  decimal  .places. 

7.  An  equilateral  triangl^is  2  in.  on  a  side.  Find  its  apothem 
and  its  area  to  two  decimal  places. 

8.  The  radius  of  one  circle  is  2^  times  that  of  another. 
The  area  of  the  smaller  is  15.2  sq.  in.  What  is  the  area  of 
the  larger  ? 

9.  The  radius  of  one  circle  is  3^  times  that  of  another. 
The  area  of  the  smaller  is  17.75  sq.  in.  What  is  the  area  of 
the  larger  ? 

10.  The  circumferences  of  two  cylindrical  steel  shafts  are 
respectively  3  in.  and  1^  in.  The  area  of  a  cross  section  of  the 
first  is  how  many  times  that  of  a  cross  section  of  the  second  ? 

11.  The  arc  of  a  sector  of  a  circle  2i  in.  in  diameter  is  1|  in. 
What  is  the  area  of  the  sector  ? 


242  BOOK  V.    PLANE  GEOMETRY 

Proposition  X.    Problem 
392.  To  inscribe  a  square  in  a  given  circle. 


Given  a  circle  with  center  O. 

Required  to  inscribe  a  square  in  the  given  circle. 

Construction.  Draw  two  diameters  A  C  and  BD  perpendicular 
to  each  other.  §  228 

Draw  AB,  BC,  CD,  and  DA. 
Then  A  BCD  is  the  square  required. 

Proof.    The  A  CBA,  DCB,  ADC,  BAD  are  rt.  A. 
{An  Z  inscribed  in  a  semicircle  is  a  rt.  Z.) 

The  A  at  the  centef  0  being  rt.  A, 
the  arcs  AB,  BC,  CD,  and  DA  are  equal, 
and  the  sides  AB,  BC,  CD,  and  DA  are  equal. 

Hence  the  quadrilateral  A  BCD  is  a  square,  by  §  65.      q.e.d. 

393.  Corollary.  To  inscribe  regular  polygons  of  ^,1Q,Z2, 
64,  etc.,  sides  in  a  given  circle. 

By  bisecting  the  arcs  AB.,  BC,  etc.,  a  regular  polygon  of  how  many 
sides  may  be  inscribed  in  the  circle  ?  By  continuing  the  process  regular 
polygons  of  how  many  sides  may  be  inscribed  ? 

In  general  we  may  say  that  this  corollary  allows  us  to  inscribe  a  reg- 
ular polygon  of  2«  sides,  where  n  is  any  positive  integer.  As  a  special 
case  it  is  interesting  to  note  that  n  may  equal  1. 


Q.E.F. 

§215 

Const. 

§212 

§170 

PEOBLEMS  OF  CONSTRUCTION      243 

Proposition  XI.    Problem 
394.   To  inscribe  a  regular  hexagon  in  a  given  circle. 


A>-~ — -^B 
Given  a  circle  with  center  0. 

Required  to  inscribe  a  regular  hexagon  in  the  given  circle. 

Construction.    From  the  center  0  draw  any  radius,  as  OC. 
With  C  as  a  center,  and  a  radius  equal  to  OC,  describe  an 
arc  intersecting  the  circle  at  D. 

Draw  OD  and  CD. 

Then  CD  is  a  side  of  the  regular  hexagon  required,  and 
therefore  the  hexagon  may  be  inscribed  by  applying  CD  six 
times  as  a  chord.  q.e.f. 

Proof.  The  A  OCD  is  equiangular.  §  75 

{An  equilateral  triangle  is  equiangular.) 

Hence  the  Z  COD  is  ^  of  2  rt.  Zs,  or  ^  of  4  rt.  A.  §  107 

.'.  the  arc  CD  is  ^  of  the  circle. 
.'.  the  chord  CD  is  a  side  of  a  regular  inscribed  hexagon,  q.  e.  d. 

395.  Corollary  1.   To  inscribe  an  equilateral  triangle  in 

a  given  circle. 

By  joining  the  alternate  vertices  of  a  regular  inscribed  hexagon,  an 
equilateral  triangle  may  be  inscribed. 

396.  Corollary  2.   To  inscribe  regular  polygons  of  12,  24, 

48,  etc.,  sides  in  a  given  circle. 


244  BOOK  V.    PLANE  GEOMETRY 

Proposition  XII.    Problem 
397.   To  inscribe  a  regular  decagon  in  a  given  circle. 


Given  a  circle  with  center  O. 

Required  to  inscribe  a  regular  decagon  in  the  given  circle. 

Construction.  Draw  any  radius  OA, 

and  divide  it  in  extreme  and  mean  ratio, 

so  that  OA:OP=OP:AP.  §  311 

From  A  as  a  center,  with  a  radius  equal  to  OP, 
describe  an  arc  intersecting  the  circle  at  B. 
Draw  AB. 

Then  A  Bis  a.  side  of  the  regular  decagon  required,  and  there- 
fore the  regular  decagon  may  be  inscribed  by  applying  AB  ten 

Q.  E.  F. 

Draw  PB  and  OB. 

OA  :  OP  =  OP  :  APj 

AB  =  OP. 

'.OA:AB  =  AB:AP. 

ZBAO  =  ZBAP. 

Hence  the  AOAB  and  BAP  are  similar. 

But  the  AOAB  is  isosceles. 

.*.  A  BAP,  which  is  similar  to  AOAB,  is  isosceles,  '§  282 

and  AB  =  BP  =  OP.  §62 


times  as  a  chord 

Proof. 

Then 
and 

Moreover, 


Const. 

Const. 
Ax.  9 
Iden. 
§288 
§162 


PROBLEMS  OF  CONSTRUCTION  .245 

The  A  PBO  being  isosceles,  the  Z  O  =  Z  OBP.  §  74 

But  the  Z^P5  =  Z0-hZ0^P  =  2Z0.  §111 

Hence  ABAP  =  2/.0, 

and                      ^            ZOi5^  =  2ZO.  Ax.  9 

.-.the  sum  of  the  A  of  the  A  OAB  =  5  Z  O  =  2  rt.  Z,  §  107 

and                   Z  O  =  1  of  2  rt.  A,  or  -^^  of  4  rt.  A.  Ax.  4 

Therefore  the  arc  ^jB  is  j^^  of  the  circle.  §  212 

.'.  the  chord  ^5  is  a  side  of  a  regular  inscribed  decagon.  q.e.d. 

398.  Corollary  1.  To  inscribe  a  regular  pentagon  in  a 
given  circle. 

399.  Corollary  2.   To  inscribe  regular  polygons  of  20,  40, 

80,  etc.^  sides  in  a  given  circle. 

By  bisecting  the  arcs  subtended  by  the  sides  of  a  regular  inscribed 
decagon  a  regular  polygon  of  how  many  sides  may  be  inscribed  ?  By  con- 
tinuing the  process  regular  polygons  of  how  many  sides  may  be  inscribed  ? 

EXERCISE  62 

If  r  denotes  the  radius  of  a  regular  inscribed  polygon,  a 
the  apothem,  s  one  side,  A  an  angle,  and  C  the  angle  at  the 
center,  show  that: 

1.  In  a  regular  inscribed  triangle  s  =  r  Vs,  a  =  ^r,  A  =  60°, 
C=120°. 

2.  In  a  regular  inscribed  quadrilateral  s  =  r  V2,  a  =  ^r  V2, 
A  =  90°,  C  =  90°. 

3.  In  a  regular  inscribed  hexagon  s  =  r,a=  ^r  Vs,  A  =  120°, 
C  =  60°. 

4.  In  a  regular  inscribed  decagon 

s  =  ^^     5-1)^  ^  ^  1^  VlO  +  2  V5,  ^  =144°,  C=  36°. 


246  BOOK  V.    PLANE  GEOMETRY 

Proposition  XIII.    Problem 

400.   To  inscribe  in  a  given  circle  a  recjulcir  jpentadec- 
agon,  or  polygon  of  fifteen  sides. 


Given  a  circle. 

Required  to  inscribe  a  regular  pentadecagon  in  the  given 
circle. 

Construction.  Draw  a  chord  PH  equal  to  the  radius  of  the 
circle,  a  chord  PA  equal  to  a  side  of  the  regular  inscribed 
decagon,  and  draw  AB. 

Then  ^5  is  a  side  of  the  regular  pentadecagon  required,  and 
therefore  the  regular  pentadecagon  may  be  inscribed  by  apply- 
ing AB  fifteen  times  as  a  chord.  q.e.f. 

Proof.  The  arc  PB  is  i  of  the  circle,  §  394 

and  the  arc  PA  is  ^^  of  the  circle.  §  397 

Hence  the  arc  AB  \?>  ^  —  J^j,  or  ^^,  of  the  circle.     Ax.  2 
.*.  the  chord  .45  is  a  side  of  the  regular  inscribed  pentadec- 
agon required.  q.e.d. 

401.  Corollary.  To  inscribe  regular  polygons  (^f  30,  60, 
120,  etc.^  sides  in  a  given  circle. 

By  bisecting  the  arcs  AB^  BC,  etc.,  a  regular  polygon  of  how  many 
sides  may  be  inscribed  ?  By  continuing  the  process  regular  polygons  of 
how  many  sides  may  be  inscribed  ?  In  general  we  may  say  that  a  regu- 
lar polygon  of  15  •  2«  sides  may  be  inscribed  in  this  manner. 


PKOBLEMS  OF  CONSTRUCTION  247 

EXERCISE  63 

1.  A  five-cent  piece  is  placed  on  the  table.  How  many  five- 
cent  pieces  can  be  placed  around  it,  each  tangent  to  it  and 
tangent  to  two  of  the  others  ?    Prove  it. 

2.  What  is  the  perimeter  of  an  equilateral  triangle  inscribed 
in  a  circle  with  radius  1  in.  ? 

3.  What  is  the  perimeter  of  an  equilateral  triangle  circum- 
scribed about  a  circle  with  radius  1  in.  ? 

4.  What  is  the  perimeter  of  a  regular  hexagon  circumscribed 
about  a  circle  with  radius  1  in,  ? 

Required  to  circumscribe  about  a  given  circle  the  following 
regular  polygons  : 

5.  Triangle.  7.  Hexagon.  9.  Pentagon. 

6.  Quadrilateral.  8.  Octagon.  10.  Decagon. 

11.  Required  to  describe  a  circle  whose  circumference  equals 
the  sum  of  the  circumferences  of  two  circles  of  given  radii. 

12.  Required  to  describe  a  circle  whose  area  equals  the  sum 
of  the  areas  of  two  circles  of  given  radii. 

13.  Required  to  describe  a  circle  having  three  times  the  area 
of  a  given  circle. 

Required  to  construct  an  angle  of  : 

14.  18°.  15.  36°.  16.  9°.  17.  12°.  18.  24°. 
Required  to  construct  tvith  a  side  of  given  length  : 

19.  An  equilateral  triangle.         23.  A  regular  pentagon. 

20.  A  square.  24.  A  regular  decagon. 

21.  A  regular  hexagon.  25.  A  regular  dodecagon. 

22.  A  regular  octagon.  26.  A  regular  pentadecagon. 
27.  From  a  circular  log  16  in.  in  diameter  a  builder  wishes 

to  cut  a  column  with  its  cross  section  as  large  a  regular  octagon 
as  possible.   Find  the  length  of  each  side. 


248  BOOK  V.    PLANE  GEOMETRY 

Proposition  XIV.    Problem 

402.  Given  the  side  and  the  radius  of  a  regular  in- 
scribed polygon,  to  find  the  side  of  the  regular  inscribed 
polygon  of  double  the  nuviber  of  sides. 


Given  AB,  the  side  of  a  regular  inscribed  polygon  of  radius  OA. 

Required  to  find  AP,  a  side  of  the  regular  inscribed  poly- 
yon  of  double  the  number  of  sides. 

Solution.  Denote  the  radius  by  r,  and  the  side  AB  by  s. 
Draw  the  diameter  PQ  1.  to  AB,  and  draw  AO  and  AQ. 
Then  AM=\s. 

In  the  rt.  A. 4 03/,   oFf  =  7^-\s^. 
Therefore  OM  =  ^r  -  i  s\ 

Since  PM-^OM=r, 

therefore  PM  =r—  OM 


Furthermore 


=  r  —  V/*^  —  1  s^. 


AP  =PQX  PM. 


But    PQ  =  2r,  and  PM  =r  —  Vr^ 


s\ 


.'.  AP  =  2  r(r  -  V?-"  -  i  s% 
.'.AP 


=  Vr(2r- V4r^-6-'). 


§174 
§338 
Ax.  5 
Ax.  11 
Ax.  2 
Ax.  9 
§298 

Ax.  9 
Ax.  5 

Q.  E.  F. 


403.  Corollary.    7/  r  =1,  ^P  =  \/2  -  V4  -  s' 


PROBLEMS  OF  COMPUTATION 
Proposition  XV.    Problem 


249 


404.   To  find  the  numerical  value  of  the  ratio  of  the 
circmnference  of  a  circle  to  its  diameter. 


Given  a  circle  of  circumference  c  and  diameter  d. 

c 
Required  to  find  the  numerical  value  of  -  or  ir. 

ct 

Solution.    By  §  385,  2  7rr  =  c.     .  • ,  tt  =  ^  c  when  r  =  1. 

Let  Sg  (read  "  s  sub  six  ")  be  the  length  of  a  side  of  a  regular 
polygon  of  6  sides,  s^^  of  12  sides,  and  so  on. 
If  r  =  1,  by  §  394,  ^<?g  =  1,  and  by  §  403  we  have 


Form  of  Computation 


=  \/2-V4-l^ 


s^  =  V2  -  V4^0.51763809)2 
s^3  =  V2  -  V4^  (0.26105238^ 
s^  =  V2  -  Vr-  (0.13080626)^ 
^m  =  V2  -  V4  -  (0.06543817)^ 
s^^  =  V2-  V4  -  (0.03272346)2 
s_g3  =  V2  -  V4^(0.01636228)^ 

.'.c  =  6.28317  nearly  ;  that  is^ 

TT  is  an  incommensurable  number.    We  generally  take 
TT  =  3.1416,  or  ^,  and  -  =  0.31831. 


Length  of  Side 

Length 
of  Perimeter 

0.51763809 

6.21165708 

0.26105238 

6.26525722 

0.13080626 

6.27870041 

0.06543817 

6.28206396 

0.03272346 

6.28290510 

0.01636228 

6.28311544 

0.00818121 

6.28316941 

=  3.14159  nearly.    Q.e.f. 

250  BOOK  V.    PLAKE  GEOMETRY 

EXERCISE  64 

Problems  of  Computation 

Using  the  value  S.1416  for  tt,  find  the  cireumferences  of 
circles  with  radii  as  follows  : 

1.  3  in.  3.  2.7  in.  5.  7|  in.  7.  2  ft.  8  in. 

2.  5  in.  4.  3.4  in.  6.   6|  in.  8.  3  ft.  7  in. 

Find  the  circumferences  of  circles  with  diameters  as  follows  : 

9.  9  in.  11.  5.9  in.  13.  2^  ft.         15.  29  centimeters. 

10.  12  in.       12.  7.3  in.  14.  3^  in.        16.  47  millimeters. 

Find  the  radii  of  circles  with  circumferences  as  follows : 

17.  Tit.         19.  15.708  in.     21.   18.8496  in.  23.  345.576  ft. 

18.  3^77.       20.  21.9912  in.  22.   125.664  in.  24.  3487.176  in. 

Find  the  diameters  of  circles  with  circumferences  as  follows  : 

25.  15  TT.       27.  2  7rr.        29.  188.496  in.       31.  3361.512  in. 

26.  7r\  28.  7  7ral       30.  219.912  in.       32.  3173.016  in. 

Find  the  areas  of  circles  ivith  radii  as  folloivs  : 

33.  5  X.         35.  27  ft.  37.  3^  in.  39.  2  ft.  6  in. 

34.  2  TT.         36.  4.8  ft.  38.  4f  in.  40.  7  ft.  9  in. 

Find  the  areas  of  circles  with  diameters  as  follows  : 

41.  16aZ».        43.2.5  ft.         45.  3|  yd.  47.  3  ft.  2  in. 

42.  24  tt".        44.  7.3  in.        46.  4f  yd.  48.  4  ft.  1  in. 

Find  the  areas  of  circles  tvith  circumferences  as  folloivs : 

49.  2  7r.       51.  ira.  53.  18.8496  in.       55.  333.0096  in. 

50.  4  7r.       52.  14  Tra^.       54.  329.868  in.       56.  364.4256  in. 

Find  the  radii  of  circles  with  areas  as  follows  : 

57.  -n-a^*.         59.  tt.  61.   12.5664.  63.  78.54. 

58.  4  7rmV.    60.  2  7r.  62.  28.2744.  64.  113.0976. 


EXEKCISES  251 

EXERCISE  65 
Problems  of  Construction 

1.  To  inscribe  in  a  given  circle  a  regular  polygon  similar  to 
a  given  regular  polygon. 

2.  To  divide  by  a  concentric  circle  the  area  of  a  given  circle 
into  two  equivalent  parts. 

3.  To  divide  by  concentric  circles  the  area  of  a  given  circle 
into  n  equivalent  parts. 

4.  To  describe  a  circle  whose  circumference  is  equal  to  the 
difference  of  two  circumferences  of  given  radii. 

5.  To  describe  a  circle  the  ratio  of  whose  area  to  that  of 
a  given  circle  shall  be  equal  to  the  given  ratio  m, :  n. 

6.  To  construct  a  regular  pentagon,  given  one  of  the 
diagonals. 

7.  To  draw  a  tangent  to  a  given  circle  such  that  the  seg- 
ment intercepted  between  the  point  of  contact  and  a  given 
line  shall  have  a  given  length. 

8.  In  a  given  equilateral  triangle  to  inscribe  three  equal 
circles  tangent  each  to  the  other  two,  each  circle  being  tangent 
to  two  sides  of  the  triangle. 

9.  In  a  given  square  to  inscribe  four  equal  circles,  so  that 
each  circle  shall  be  tangent  to  two  of  the  others  and  also 
tangent  to  two  sides  of  the  square. 

10.  In  a  given  square  to  inscribe  four  equal  circles,  so  that 
each  circle  shall  be  tangent  to  two  of  the  others  and  also 
tangent  to  one  side  and  only  one  side  of  the  square. 

11.  To  draw  a  common  secant  to  two  given  circles  exterior 
to  each  other,  such  that  the  intercepted  chords  shall  have  the 
given  lengths  a  and  b. 

12.  To  draw  through  a  point  of  intersection  of  two  given 
intersecting  circles  a  common  secant  of  a  given  length. 


252  BOOK  V.    PLANE  GEOMETRY 

EXERCISE  66 
Problems  of  Loci 

1.  Find  the  locus  of  the  center  of  the  circle  inscribed  in  a 
triangle  that  has  a  given  base  and  a  given  angle  at  the  vertex. 

2.  Find  the  locus  of  the  intersection  of  the  perpendiculars 
from  the  three  vertices  to  the  opposite  sides  of  a  triangle  that 
has  a  given  base  and  a  given  angle  at  the  vertex. 

3.  Find  the  locus  of  the  extremity  of  a  tangent  to  a  given 
circle,  if  the  length  of  the  tangent  is  equal  to  a  given  line. 

4.  Find  the  locus  of  a  point  from  which  tangents  drawn 
to  a  given  circle  form  a  given  angle. 

5.  Find  the  locus  of  the  mid-point  of  a  line  drawn  from 
a  given  point  to  a  given  line. 

6.  Find  the  locus  of  the  vertex  of  a  triangle  that  has  a 
given  base  and  a  given  altitude. 

7.  Find  the  locus  of  a  point  the  sum  of  whose  distances 
from  two  given  parallel  lines  is  constant. 

8.  Find  the  locus  of  a  point  the  difference  of  whose  dis- 
tances from  two  given  parallel  lines  is  constant. 

9.  Find  the  locus  of  a  point  the  sum  of  whose  distances 
from  two  given  intersecting  lines  is  constant. 

10.  Find  the  locus  of  a  point  the  difference  of  whose  dis- 
tances from  two  given  intersecting  lines  is  constant. 

11.  Find  the  locus  of  a  point  whose  distances  from  two 
given  points  are  in  the  ratio  m  :  n. 

12.  Find  the  locus  of  a  point  whose  distances  from  two 
given  parallel  lines  are  in  the  ratio  m  :  n. 

13.  Find  the  locus  of  a  point  whose  distances  from  two 
given  intersecting  lines  are  in  the  ratio  m:n. 

14.  Find  the  locus  of  a  point  the  sum  of  the  squares  of 
whose  distances  from  two  given  points  is  constant. 


EXERCISES  253 

EXERCISE  67 
Examination  Questions 

1.  Each  side  of  a  triangle  is  2  ti  centimeters,  and  about  each 
vertex  as  a  center,  with  a  radius  of  n  centimeters,  a  circle  is 
described.  Find  the  area  bounded  by  the  three  arcs  that  lie 
outside  the  trid:ngle,  and  the  area  bounded  by  the  three  arcs 
that  lie  within  the  triangle. 

2.  Upon  a  line  AB  2i  segment  of  a  circle  containing  240°  is 
constructed,  and  in  the  segment  any  chord  CD  subtending  an 
arc  of  60°  is  drawn.  Find  the  locus  of  the  intersection  of  ^C 
and  J5Z),  and  also  of  the  intersection  oi  AD  and  BC. 

3.  Three  successive  vertices  of  a  regular  octagon  are  A,  B, 
and  C.    If  the  length  AB  is  a,  compute  the  length  A C. 

4.  The  areas  of  similar  segments  of  circles  are  proportional 
to  the  squares  on  their  radii. 

5.  An  arc  of  a  certain  circle  is  100  ft.  long  and  subtends 
an  angle  of  25°  at  the  center.  Compute  the  radius  of  the  circle 
correct  to  one  decimal  place. 

6.  Given  a  circle  whose  radius  is  16,  find  the  perimeter  and 
the  area  of  the  regular  inscribed  octagon. 

7.  If  two  circles  intersect  at  the  points  A  and  B,  and  through 
A  a  variable  secant  is  drawn,  cutting  the  circles  in  C  and  D,  the 
angle  CBD  is  constant  for  all  positions  of  the  secant. 

8.  If  A  and  B  are  two  fixed  points  on  a  given  circle,  and  P 
and  Q  are  the  extremities  of  a  variable  diameter  of  the  same 
circle,  find  the  locus  of  the  point  of  intersection  of  the  lines 
AP  and  BQ. 

9.  The  radius  of  a  circle  is  10  ft.  Two  parallel  chords  are 
drawn,  each  equal  to  the  radius.  Find  that  part  of  the  area  of 
the  circle  lying  between  the  parallel  chords. 

The  propositions  in  Exercise  67  are  taken  from  recent  college  entrance 
examination  papers. 


254  BOOK  V.    PLANE  GEOMETRY 

EXERCISE  68 

Formulas 

If  r  denotes  the  radius  of  a  circle^  and  s  one  side  of  a  reg- 
ular inscribed  polygon,  prove  the  following,  and  find  the  value 
of  s  to  two  decimal  places  when  r  —1 : 

1.  In  an  equilateral  triangle  s  =  r  VS. 

2.  In  a  square  s  =  r  V2. 

3.  In  a  regular  pentagon  5  =  ^  ?•  V 10  —  2  V 5. 

4.  In  a  regular  hexagon  s  =  r. 

5.  In  a  regular  octagon  s  =  r  v2  —  V2. 

6.  In  a  regular  decagon  s  =  ^  r (Vs  —  1). 


7.  In  a  regular  dodecagon  s  =  r\2  —  V 3. 

8.  A  regular  pentagon  is  inscribed  in  a  circle  whose  radius 
is  r.    If  the  side  is  s,  find  the  apothem. 

9.  A  regular  polygon  is  inscribed  in  a  circle  whose  radius 
is  r.    If  the  side  is  s,  show  that  the  apothem  is  \  V4  7^  —  s\ 

10.  If  the  radius  of  a  circle  is  r,  and  the  side  of  an  inscribed 
regular  polygon  is  s,  show  that  the  side  of  the  similar  cir- 
cumscribed regular  polygon  is 


11.  Three  equal  circles  are  described,  each  tangent  to  the 
other  two.  If  the  common  radius  is  r,  find  the  area  contained 
between  the  circles. 

12.  Given  p,  P,  the  perimeters  of  regular  polygons  of  n  sides 
inscribed  in  and  circumscribed  about  a  given  circle.  Eind  p', 
P',  the  perimeters  of  regular  polygons  of  2n  sides  inscribed  in 
and  circumscribed  about  the  given  circle. 

13.  A  circular  plot  of  land  d  ft.  in  diameter  is  surrounded 
by  a  walk  w  ft.  wide.  Find  the  area  of  the  circular  plot  and 
the  area  of  the  walk. 


EXERCISES  255 

EXERCISE  69 
Applied  Problems 

1.  The  diameter  of  a  bicycle  wheel  is  28  in.  How  many 
revolutions  does  the  wheel  make  in  going  10  mi.  ? 

2.  Find  the  diameter  of  a  carriage  wheel  that  makes  264 
revolutions  in  going  half  a  mile. 

3.  A  circular  pond  100  yd.  in  diameter  is  surrounded  by  a 
walk  10  ft.  wide.    Find  the  area  of  the  walk. 

4.  The  span  (chord)  of  a  bridge  in  the  form  of  a  circular 
arc  is  120  ft.,  and  the  highest  point  of  the  arch  is  15  ft.  above 
the  piers.    Find  the  radius  of  the  arc. 

5.  Two  branch  water  pipes  lead  into  a  main  pipe.  It  is 
necessary  that  the  cross-section  area  of  the  main  pipe  shall 
equal  the  sum  of  the  cross  sections  of  the  two  branch  pipes. 
The  diameters  of  the  branch  pipes  are  respectively  3  in.  and 
4  in.    Required  the  diameter  of  the  main  pipe.  

6.  A  kite  is  made  as  here  shown,  the  semicircle   /         ,   \ 
having  a  radius  of  9  in.,  and  the  triangle  a  height 
of  25  in.    Find  the  area  of  the  kite. 

7.  In  making  a  drawing  for  an  arch  it  is  required 
to  mark  off  on  a  circle  drawn  with  a  radius  of  5  in. 
an  arc  that  shall  be  8  in.  long.    This  is  best  done  by 

finding  the  angle  at  the  center.    How  many  degrees  are  there  in 
this  angle  ? 

8.  In  an  iron  washer  here  shown,  the  diameter 
of  the  hole  is  1|  in.  and  the  width  of  the  washer 
is  I  in.    Find  the  area  of  one  face  of  the  washer. 

9.  Find  the  area  of  a  fan  that  opens  out  into  a  sector  of 
120°,  the  radius  being  9|  in. 

10.  The  area  of  a  fan  that  opens  out  into  a  sector  of  111°  is 
96.866  sq.  in.    What  is  the  radius  ?    (Take  tt  =  3.1416.) 


256  BOOK  V.    PLANE  GEOMETRY 

EXERCISE  70 
Review  Questions 

1.  What  is  meant  by  a  regular  polygon  ?  by  its  radius  ? 
by  its  center?  by  its  apothem  ? 

2.  What  other  names  are  there  for  a  regular  triangle  and 
a  regular  quadrilateral  ? 

3.  If  one  angle  of  a  regular  polygon  is  known,  how  can 
the  number  of  sides  be  determined? 

4.  The  sides  of  two  regular  polygons  of  n  sides  are  respec- 
tively s  and  s\  What  is  the  ratio  of  their  radii  ?  of  their 
apothems  ?  of  their  perimeters  ?  of  their  areas  ? 

5.  The  diameters  of  two  circles  are  d  and  d'  respectively. 
What  is  the  ratio  of  their  radii  ?  of  their  circumferences  ?  of 
their  areas  ? 

6.  If  the  number  of  sides  of  a  regular  inscribed  polygon 
is  indefinitely  increased,  what  is  the  limit  of  the  apothem? 
of  each  side  ?  of  the  perimeter  ?  of  the  area  ?  of  the  angle 
at  the  center  ?  of  each  angle  of  the  polygon  ? 

7.  How  do  you  find  the  area  of  a  regular  polygon  ?  of  an 
irregular  polygon  ?  of  a  square  ?  of  a  triangle  ?  of  a  parallelo- 
gram ?  of  a  circle  ?  of  a  trapezoid  ?  of  a  sector  ? 

8.  What  regular  polygons  have  you  learned  to  inscribe  in 
a  circle  ?  Name  three  regular  polygons  that  you  have  not 
learned  to  inscribe. 

9.  Given  the  circumference  of  a  circle,  how  can  the  area  of 
the  circle  be  found  ? 

10.  Given  the  area  of  a  circle,  how  can  the  circumference  of 
the  circle  be  found  ? 

11.  What  is  the  radius  of  the  circle  of  which  the  number  of 
linear  units  of  circumference  is  equal  to  the  number  of  square 
units  of  area  ? 


EXEECISES  257 

EXERCISE  71 
General  Review  of  Plane  Geometry 

Write  a  classification  of  the  different  kinds  of: 

1.  Lines.  3.  Triangles.  5.  Polygons. 

2.  Angles.  4.  Quadrilaterals.  6.  Parallelograms. 

State  the  conditions  under  which : 

7.  Two  triangles  are  congruent. 

8.  Two  parallelograms  are  congruent. 

9.  Two  triangles  are  similar. 

10.  Two  straight  lines  are  parallel. 

11.  Two  parallelograms  are  equivalent. 

12.  Two  polygons  are  similar. 

Complete  the  following  statements  in  the  most  general  manner: 

13.  In  any  triangle  the  square  on  the  side  opposite  •  •  •. 

14.  If  two  parallel  lines  are  cut  by  a  transversal,  •  •  •. 

15.  If  four  quantities  are  in  proportion,  they  are  in  pro- 
portion by  •  •  • . 

16.  If  two  secants  of  a  circle  intersect,  the  angle  formed  is 
measured  by  •••. 

17.  The  perimeters  of   two  similar  polygons  are   to  each 
other  as  •  •  • . 

18.  The  areas  of  two  similar  polygons  are  to  each  other  as  •  •  •. 

19.  The  area  of  a  circle  is  equal  to  •  •  •. 

20.  In  the  same  circle  or  in  equal  circles  equal  chords  •  •  •. 

21.  In  the  same  circle  or  in  equal  circles  the  central  angles 
subtended  by  two  arcs  are  •  •  • . 

22.  If  two  secants  of  a  circle  intersect  within,  on,  or  outside 
the  circle,  the  product  of  •  •  • . 


258  BOOK  V.    PLANE  GEOMETRY 

23.  If  four  lines  meet  in  a  point  so  that  the  opposite  angles 
are  equal,  these  lines  form  two  intersecting  straight  lines. 

24.  If  squares  are  constructed  outwardly  on  the  six  sides 
of  a  regular  hexagon,  the  exterior  vertices  of  these  squares  are 
the  vertices  of  a  regular  dodecagon. 

25.  In  a  right  triangle  the  line  joining  the  vertex  of  the 
right  angle  to  the  mid-point  of  the  hypotenuse  is  equal  to  half 
the  hypotenuse. 

26.  No  two  lines  drawn  from  the  vertices  of  the  base  angles 
of  a  triangle  to  the  opposite  sides  can  bisect  each  other. 

27.  The  rhombus  is  the  only  parallelogram  that  can  be  cir- 
cumscribed about  a  circle. 

28.  The  square  is  the  only  rectangle  that  can  be  circum- 
scribed about  a  circle. 

29.  No  oblique  parallelogram  can  be  inscribed  in  a  circle. 

30.  If  two  triangles  have  equal  bases  and  equal  vertical 
angles,  the  two  circumscribing  circles  have  equal  diameters. 

31.  If  the  inscribed  and  circumscribed  circles  of  a  triangle 
are  concentric,  the  triangle  is  equilateral. 

32.  If  the  three  points  of  contact  of  a  circle  inscribed  in  a  tri- 
angle are  joined,  the  angles  of  the  resulting  triangle  are  all  acute. 

33.  The  diagonals  of  a  regular  pentagon  intersect  at  the 
vertices  of  another  regular  pentagon. 

34.  If  two  perpendicular  radii  of  a  circle  are  produced  to 
intersect  a  tangent  to  the  circle,  the  other  tangents  from  the 
two  points  of  intersection  are  parallel. 

35.  The  line  that  joins  the  feet  of  the  perpendiculars  drawn 
from  the  extremities  of  the  base  of  an  isosceles  triangle  to  the 
equal  sides  is  parallel  to  the  base. 

36.  The  sum  of  the  perpendiculars  drawn  to  the  sides  of  a 
regular  polygon  from  any  point  within  the  polygon  is  equal 
to  the  apothem  multiplied  by  the  number  of  sides. 


EXERCISES  259 

37.  If  two  consecutive  angles  of  a  quadrilateral  are  right 
angles,  the  bisectors  of  the  other  two  angles  are  perpendicular. 

38.  If  two  opposite  angles  of  a  quadrilateral  are  right  angles, 
the  bisectors  of  the  other  two  angles  are  parallel. 

39.  The  two  lines  that  join  the  mid-points  of  opposite  sides 
of  a  quadrilateral  bisect  each  other. 

40.  The  sum  of  the  angles  at  the  vertices  of  a  five-pointed 
star  is  equal  to  two  right  angles. 

41.  The  segments  of  any  line  intercepted  between  two  con- 
centric circles  are  equal. 

42.  The  diagonals  of  a  trapezoid  divide  each  other  into 
segments  which  are  proportional. 

43.  Given  the  mid-points  of  the  sides  of  a  triangle,  to  con- 
struct the  triangle. 

44.  To  divide  a  given  triangle  into  two  equivalent  parts  by 
a  line  through  one  of  the  vertices. 

45.  To  draw  a  tangent  to  a  given  circle  that  shall  also  be 
perpendicular  to  a  given  line. 

46.  To  divide  a  given  line  into  two  segments  such  that  the 
square  on  one  shall  be  double  the  square  on  the  other. 

47.  If  any  two  consecutive  sides  of  an  inscribed  hexagon 
are  respectively  parallel  to  their  opposite  sides,  the  remaining 
two  sides  are  parallel. 

48.  If  through  any  given  point  in  the  common  chord  of  two 
intersecting  circles  two  other  chords  are  drawn,  one  in  each 
circle,  their  four  extremities  will  all  lie  on  a  third  circle. 

49.  If  two  chords  intersect  at  right  angles  within  a  circle, 
the  sum  of  the  squares  on  their  segments  equals  the  square 
on  the  diameter.  Investigate  the  case  in  which  the  chords 
intersect  outside  the  circle ;  also  the  case  in  which  they  inter- 
sect on  the  circle. 


260  BOOK  V.    PLANE  GEOMETRY 

50.  The  lines  bisecting  any  angle  of  an  inscribed  quadrilateral 
and  the  opposite  exterior  angle  intersect  on  the  circle. 

51.  The  sum  of  the  perpendiculars  from  any  point  in  an 
equilateral  triangle  to  the  three  sides  is  constant. 

52.  The  perpendiculars  from  the  vertices  of  a  triangle  upon 
the  opposite  sides  cut  one  another  into  segments  that  are 
reciprocally  proportional  to  each  other. 

53.  The  area  of  a  triangle  is  equal  to  half  the  product  of  its 
perimeter  by  the  radius  of  the  inscribed  circle. 

54.  The  perimeter  of  a  triangle  is  to  one  side  as  the  perpen- 
dicular from  the  opposite  vertex  is  to  the  radius  of  the  inscribed 
circle. 

55.  The  area  of  a  square  inscribed  in  a  semicircle  is  equal  to 
two  fifths  of  the  area  of  the  square  inscribed  in  the  circle. 

56.  The  diagonals  of  any  inscribed  quadrilateral  divide  it 
into  two  pairs  of  similar  triangles. 

57.  To  draw  a  line  whose  length  is  Vt^  in. 

58.  If  two  equivalent  triangles  are  on  the  same  base  and  the 
same  side  of  the  base,  any  line  cutting  the  triangles,  and  par- 
allel to  the  base,  cuts  off  equal  areas  from  the  triangles. 

59.  To  divide  a  given  arc  of  a  circle  into  two  parts  such  that 
their  chords  shall  be  in  a  given  ratio. 

60.  The  areas  between  two  concentric  circles  may  be  found 
by  multiplying  half  the  sum  of  the  two  circumferences  by  the 
difference  between  the  radii. 

61.  Find  the  length  of  the  belt  connecting  two  wheels  of 
the  same  size,  if  the  radius  of  each  wheel  is  18  in.,  the  distance 
between  the  centers  6  ft.,  and  4  in.  is  allowed  for  sagging. 

62.  To  construct  a  regular  inscribed  heptagon  draftsmen 
sometimes  use  for  a  side  half  the  side  of  an  inscribed  equi- 
lateral triangle.  Construct  such  a  figure  with  the  compasses, 
and  state  whether  the  rule  seems  exact  or  only  approximate. 


APPENDIX 


405.  Subjects  Treated.  Of  the  many  additionar  subjects  that 
may  occupy  the  attention  of  the  student  of  plane  geometry  if 
time  permits,  two  are  of  special  interest.  These  are  Symmetry, 
and  Maxima  and  Minima. 

406.  Symmetric  Points.  Two  points  are  said  to  be  symmetric 
with  respect  to  a  p)oi7it,  called  the  center  of  symmetry,  if  this 
third  point  bisects  the  straight  line  which  joins  the  two  points. 

Two  points  are  said  to  be  symmetric  with  respect  to  an  axis, 
if  a  straight  line,  called  the  axis  of  symmetry,  is  the  perpen- 
dicular bisector  of  the  line  joining  them. 

407.  Symmetric  Figure.  A  figure  is  said 
to  be  symmetric  with  resjject  to  a  pjoint,  if 
the  point  bisects  every  straight  line  drawn 
through  it  and  terminated  by  the  boundary 
of  the  figure. 

A  figure  is  said  to  be  symmetric  with 
respect  to  an  axis,  if  the  axis  bisects  every      /_ 
perpendicular  through  it  and  terminated 
by  the  boundary  of  the  figure. 

Evidently  this  will  be  the  case  if  one  part  coin- 
cides with  another  part  when  folded  over  the  axis. 

408.  Two  Symmetric  Figures.  Two  figures 
are  said  to  be  symmetric  ivith  respect  to  a 
point  or  symmetric  with  respect  to  an  axis, 
if  every  point  of  each  has  a  correspond- 
ing symmetric  point  in  the  other. 

261 


B' 


262 


APPENDIX  TO  PLANE  GEOMETRY 


Pkoposition  I.    Theorem 

A  quadrilateral  that  has  tioo  adjacent  sides 
and  the  other  tivo  sides  equal,  is  symmetric 
loith  respect  to  the  diagonal  joining  the  vertices  of  the 
angles  formed  hy  the  equal  sides;  and  the  diagonals 
are  perpendicular  to  each  other. 


409 

equal 


Given  the  quadrilateral  ABCD,  having  AB  equal  to  AD,  and  CB 
equal  to  CD,  and  having  the  diagonals  AC  and  BD. 

To  prove  that  the  diagonal  A  C  is  an  axis  of  symmetry^  and 
that  AC  is  A.  to  BD. 

Proof.  In  the  A  yl/?C  and  ADC, 

AB  =  AD,  and  CB  =  CD,  Given 

and                                         AC  =  AC.  Iden. 

.'.  A  ABC  is  congruent  to  A  ADC.  §  80 

.•.ZBAC  =  ZCAD,  and  ZACB=:ZDCA.  §67 

Hence,  if  A  ^5C  is  turned  on  ^C  as  an  axis  until  it  falls 
on  A  ADC,  AB  will  fall  on  AD,  CB  on  CD,  and  OB  on  OD. 
.-.  the  A  ABC  will  coincide  with  the  A  ADC. 
.'.AC  will  bisect  every  perpendicular  drawn  through  it  and 
terminated  by  the  boundary  of  the  figure. 

.*.  ^C  is  an  axis  of  symmetry.  §  407 

.-.  ^C  is  ±  to  BD,  by  §  406.  Q.e.d. 


SYMMETRY 


263 


Proposition  II.    Theorem 

410.  If  a  figure  is  symmetric  with  respect  to  tioo 
axes  j^erpendicular  to  each  other,  it  is  symmetric  with 
respect  to  their  intersection  as  a  center. 


- 
C 

B 

L 

A 

E 

0 

G 

Y' 

Given  the  figure  ABCDEFGH,  symmetric.with  respect  to  the  two 
perpendicular  axes  XX^,  YY',  which  intersect  at  O. 

To  prove  that  0  is  the  center  of  symmetry  of  the  fiyure. 
Proof.  Let  P  be  any  point  in  the  perimeter. 

Draw  PMQ  J_  to  YY\  and  QNR  _L  to  XX\ 
Then  PQ  is  II  to  XX\  and  QR  is  II  to  YY\ 

Draw  PO,  OR,  and  MN. 
Then  QN  =  NR. 

{The figure  is  given  as  symmetric  with  respect  to  XX \) 

But  QN  =  MO. 

.\NR  =  MO. 

.'.  RO  is  equal  and  parallel  to  NM. 
In  like  manner,  OP  is  equal  and  parallel  to  NM. 
.'.  ROP  is  a  straight  line. 

.'.0  bisects  PR,  any  straight  line,  and  hence  bisects  every 

straight  line  drawn  through  0  and  terminated  by  the  perimeter. 

.*.  O  is  the  center  of  symmetry  of  the  figure,  by  §  407.    Q-E.d. 


§227 
§95 

§407 

§127 
Ax.  8 
§130 

§94 


264  APPENDIX  TO  PLANE  GEOMETRY 

EXERCISE  72 

1.  Draw  a  figure  showing  the  number  of  axes  of  symmetry- 
possessed  by  a  square. 

2.  Draw  a  figure  showing  the  number  of  axes  of  symmetry 
possessed  by  a  regular  hexagon. 

3.  Draw  a  figure  showing  six  of  the  unlimited  number  of  axes 
of  symmetry  of  a  circle,  and  showing  the  center  of  symmetry. 

4.  Show  by  drawings  that  two  congruent  triangles  may 
be  placed  in  a  position  of  symmetry  with  respect  to  an  axis. 
In  one  of  the  drawings  let  a  common  side  be  the  axis. 

5.  Show  by  a  drawing  that  two  congruent  triangles  may  be 
placed  in  a  position  of  symmetry  with  respect  to  a  center. 

6.  Two  figures  symmetric  with  respect  to  an  axis  are  con- 
gruent. 

7.  Two  figures  symmetric  with  respect  to  a  center  are  con- 
gruent. 

8.  Make  a  list  of  quadrilaterals  that  are  symmetric  with 
respect  to  an  axis. 

9.  Make  a  list  of  quadrilaterals  that  are  symmetric  with 
respect  to  a  center. 

10.  What  kinds  of  regular  polygons  are  symmetric  with  re- 
spect both  to  a  center  and  to  an  axis  ?  Prove  this  for  the  hexagon. 

11.  A  circle  is  symmetric  with  respect  to  its  center  as  a 
center  of  symmetry,  and  is  also  symmetric  with  respect  to 
any  diameter  as  an  axis. 

12.  An  isosceles  triangle  is  symmetric  with  respect  to  an  axis, 
and  therefore  the  angles  opposite  the  equal  sides  are  equal. 

13.  Two  tangents  drawn  to  a  circle  from  the  same  point  are 
symmetric  with  respect  to  an  axis. 

14.  The  four  common  tangents  to  two  given  circles  form, 
together  with  the  circles,  a  figure  symmetric  with  respect  to 
the  line  of  centers  as  an  axis. 


MAXIMA  AND  MINIMA 


265 


411.  Maxima  and  Minima.  Among  geometric  magnitudes  that 
satisfy  given  conditions,  the  greatest  is  called  the  maximum,  and 
the  smallest  is  called  the  minimum. 

The  plural  of  maximum  is  maxima,  and  the  plural  of  minimum  is 
minima. 

Among  geometric  magnitudes  that  satisfy  given  conditions,  there  may 
be  several  equal  magnitudes  that  are  greater  than  any  others.  In  this 
case  all  are  called  maxima. . 

Similarly  there  may  be  several  minima  magnitudes  of  a  given  kind. 

412.  Isoperimetric  Polygons.  Polygons  which  have  equal 
perimeters  are  called  isoperimetric  polygons. 

If  the  circumference  of  a  circle  equals  the  perimeter  of  a  polygon,  the 
circle  and  the  polygon  are  said  to  be  isoperimetric,  and  similarly  for  all 
other  closed  figures  in  a  plane. 


Proposition  III.    Theorem 

413.   Of  all  triangles  having  two  given  sides,  that  in 
which  these  sides  include  a  right  angle  is  the  maximum. 


P  A  F  B 

Given  the  triangles  ABC  and  ABDy  with  AB  and  CA  equal  to 
AB  and  DA  respectively,  and  with  angle  BAC  a  right  angle. 

To  prove  that        AABC>  A  ABD. 

Proof.               Prom  D  draw  the  altitude  DP.  §  227 

Then                               DA  >  DP.  §  86 

But                                 DA  =  CA.  Given 

.-.  CA  >  DP.  Ax.  9 

.'.A  ABC  >AABD,  by  §  327.  Q.e.d. 


266  APPENDIX  TO  PLANE  GEOMETRY 

Proposition  IV.    Theorem 

414.   Of  all  isoj^erimetric  triangles  having  the  same 
base  the  isosceles  triangle  is  the  maximum. 


B' 

yi 

./' 

'/> 

/ 

^;^ 

/h" 

/   1 

/ 

-HQ 

Fig.  1 


Fig.  2 


§135 

§215 

§97 


Given  the  triangles  ABC  and  ABC  having  equal  perimeters,  and 
having  AC  equal  to  -BC,  and  AC  not  equal  to  BC\ 

To  prove  that         A  ABC  >  A  ABC. 

Proof.         Produce  AC  to  B',  making  CB'  =  AC. 

Draw  BB'  and  C'B',  and  draw  C(2  II  to  AB. 
Then  since  AC  =  CB',  .\  BQ  =  QB'. 

And  since    CA  =  CB  =  CB',  .-.ZB 'BA  is  a  rt.  Z. 
.-.  CQ  is  _L  to  BB'. 

C  cannot  lie  on  AB',  for  if  it  could,  then  CC'+C'B  would 

equal  CB,  which  is  impossible.  Post.  1 

Then  since             AC -i-CB' <AC' -\- C'B',  §112 

.-.  AC-\-CB<AC'-\-C'B'.  Ax.  9 

.'.  AC'-\-C'B<AC'-{-C'B'.  Ax.  9 

.-.  C'B <  C'B'.  Ax.  6 

.-.  C '  cannot  lie  on  CQ,  for  then  C  'B  would  equal  C  'B'.   §  150 

C'cannot  lie  above  CQ  (Fig.  1),  for  C"i>",  which  <  C'P  +  P^', 
would  be  less  than  C'B,  which  equals  C'P  +  PB'. 

.'.  C  must  lie  below  CQ,  as  in  Eig.  2. 

.-.  A  ABC  >  A  ABC,  by  §  327.  Q.e.d. 


MAXIMA  AND  MINIMA  267 

Proposition  V.    Theorem 

415.  Of  all  polygons  with  sides  all  given  hut  one,  the 
maximum  can  he  inscribed  in  the  semicircle  ivhich  has 
the  undetermined  side  for  its  diameter. 

c 


Given  ABCDE^  the  maximum  of  polygons  with  sides  AB^  BC^ 
CDj  DEj  having  the  vertices  A  and  E  on  the  line  MN. 

To  prove  that  ABODE  can  he  inscribed  in  the  semicircle 
having  EA  for  its  diameter. 

Proof.    From  any  vertex,  as  C,  draw  CA  and  CE. 

The  A  ACE  must  be  the  maximum  of  all  A  having  the  sides 
CA  and  CE,  and  the  third  side  on  MN ;  otherwise,  by  increas- 
ing or  diminishing  the  Z  EC  A,  keeping  the  lengths  of  the  sides 
CA  and  CE  unchanged,  but  sliding  the  extremities  A  and  E 
along  the  line  MN,  we  could  increase  the  A  A  CE,  while  the 
rest  of  the  polygon  would  remain  unchanged;  and  therefore 
we  could  increase  the  polygon.  But  this  is  contrary  to  the 
hypothesis  that  the  polygon  is  the  maximum  polygon. 

Hence  the  A  ACE  is  the  maximum  of  triangles  that  have 
the  sides  CA  and  CE. 

Therefore  the  Z.  A  CE  is  a  right  angle.  §  413 

Therefore  C  lies  on  the  semicircle  having  EA  for  its 
diameter.  §  215 

Hence  every  vertex  lies  on  this  semicircle. 

That  is,  the  maximum  polygon  can  be  inscribed  in  the  semi- 
circle having  the  undetermined  side  for  its  diameter.         q.e.d. 


268 


APPENDIX  TO  PLANE  GEOMETRY 


Proposition  VI.    Theorem 

416.    Of  all  polygons  ivitJi  given  sides,  one  that  can 
he  inscribed  in  a  circle  is  the  maximum. 


Given  the  polygon  ABCDE  inscribed  in  a  circle,  and  the  polygon 
A^B'C'D'E'  which  has  its  sides  equal  respectively  to  the  sides  of 
ABCDE ^  but  which  cannot  be  inscribed  in  a  circle. 

To  prove  that       ABCDE  >  A'B'C'B'U'. 

Proof.    Draw  the  diameter  AP^  and  draw  CP  and  PD. 

Upon  CD'  as  a  base,  construct  the  A  C'P'D'  congruent  to  the 
ACPD,  and  draw  ^'P'. 

Since,  by  hypothesis,  a  O  cannot  pass  through  all  the  vertices 
of  A'B'C'P'D'E',  one  or  both  of  the  parts  A'P'D'E',  A'B'C'P' 
cannot  be  inscribed  in  a  semicircle. 

Neither  A'P'D'E'  or  A'B'C'P'  can  be  greater  than  its  corre- 
sponding part.  §  415 

{Of  all  polygons  with  sides  all  given  but  one,  the  maximum  can  be  inscribed 
in  the  semicircle  which  Jias  the  undetermined  side  for  its  diameter.) 

Therefore  one  of  the  parts  A'P'D'E',  A'B'C'P'  must  be  less 
than,  and  the  other  cannot  be  greater  than,  the  corresponding 
part  of  ABCPDE. 

.-.  ABC PDE>  A'B'C'P'D'E'. 

Take  from  the  two  figures  the  congruent  A  CPD  and  C'P'B'. 

Then  ABCDE  >A'B'C'D'E',  by  Ax.  6.  Q.e.d. 


MAXIMA  AND  MINIMA 
Proposition  VII.    Theorem 


269 


417.    Of  isoperwietric  polygons  of  a  given  niiinhe?'  of 
sides,  the  maximum  is  equilateral. 


Given  the  polygon  ABCDEF,  the  maximum  of  isoperimetric 
polygons  of  n  sides. 

To  prove  that  the  polygon  ABCDEF  is  equilateral. 
Proof.  Draw  A  C. 

The  A  ABC  must  be  the  maximum  of  all  the  A  which  are 
formed  upon  AC  with  a  perimeter  equal  to  that  of  A  ABC. 

Otherwise  a  greater  A  A  PC  could  be  substituted  for  AABC^ 
without  changing  the  perimeter  of  the  polygon. 

But  this  is  inconsistent  with  the  fact  that  the  polygon 
ABCDEF  is  given  as  the  maximum  polygon. 

.*.  the  A  ABC  is  isosceles.  §  414 

.\AB  =  BC. 
Similarly         BC  =  CD,  CD  =  DE,  and  so  on. 

.'.  the  polygon  ABCDEF  is  equilateral.  q.e.d. 

418.  Corollary.  The  maximum  of  isoperimetric  polygons 
of  a  given  number  of  sides  is  a  regular  polygon. 

For  the  maximum  polygon  is  equilateral  (§  417),  and  can  be  inscribed 
in  a  circle  (§  416).   Therefore  the  maximum  polygon  is  regular  (§  365). 


270  APPENDIX  TO  PLANE  GEOMETRY 

Proposition  VIII.    Theorem 

419.   Of  isoperwietric  regular  polygons,  that  ivhich 
has  the  greatest  nimiber  of  sides  is  the  maximimi. 


Y< 


Given  the  regular  polygon  P  of  three  sides,  and  the  isoperimetric 
regular  polygon  P'  of  four  sides. 

To  prove  that  P'>P. 

Proof.   Draw  CX  from  C  to  any  point  X  in  ^jB. 

Invert  the  A  AXC  and  place  it  in  the  position  XCY,  letting 
X  fall  at  C,  C  at  X,  and  yl  at  F. 

The  polygon  XBCY  is  an  irregular  polygon  of  four  sides, 
which  by  construction  has  the  same  perimeter  as  P'  and  the 
same  area  as  P. 

Then  the  regular  polygon  P'  of  four  sides  is  greater  than 
the  isoperimetric  irregular  polygon  XBCY  of  four  sides.  §  418 

That  is,  a  regular  polygon  of  four  sides  is  greater  than  the 
isoperimetric  regular  polygon  of  three  sides. 

In  like  manner,  it  may  be  shown  that  P '  is  less  than  the  iso- 
perimetric regular  polygon  of  five  sides,  and  so  on.  q.e.d. 

Discussion.  We  may  illustrate  this  by  the  case  of  an  equilateral  tri- 
angle and  a  square,  eacji  with  the  perimeter.^.  In  the  triangle  the  base 
is  I  p,  the  altitude  I  p  Vs,  and  the  area  J^p2  Vs,  or  about  0.048  p^.  In  the 
square  the  base  and  altitude  are  each  i  p,  and  the  area  is  ^^  p^,  or  0.0625  jp^. 
The  area  of  the  polygon  is  therefore  increasing  as  we  increase  the  number 
of  sides. 

Since  the  limit  approached  by  the  perimeters  is  a  circle,  we  may  infer 
that  of  all  isoperimetric  plane  figures  the  circle  has' the  greatest  area. 


MAXIMA  AND  MINIMA 


271 


Proposition  IX.    Theorem 

420.   Of  regular  polygons  having  a  given  area,  that 
which  has  the  greatest  nuviber  of  sides  has  the  minimum 

2Jerimeter. 


Given  the  regular  polygons  P  and  P'  having  the  same  area,  P' 
having  the  greater  number  of  sides. 

To  prove  that  the  perimeter  of  P  >  the  perimeter  of  P'. 

Proof.    Construct  the  regular  polygon  P"  having  the  same 
perimeter  as  P  \  and  the  same  number  of  sides  as  P. 

Denote  a  side  of  P  by  s,  and  a  side  of  P  "  by  s". 


§419 

Given 

Ax.  9 

§374 


Then  P'>P". 

But  P  =  P'. 

.-.  P>P". 
But  P:P"  =  s^:s"\ 

.\s>s".  Ax.  6 

.'.  the  perimeter  of  P  >  the  perimeter  of  P".       Ax.  6 

.  But  the  perimeter  of  P '  =  the  perimeter  of  P  ".    Const. 

.'.  the  perimeter  of  P  >  the  perimeter  of  P',  by  Ax.  9.    q.e.  d. 

Discussion.  We  may  illustrate  this,  as  on  page  270,  by  the  case  of 
an  equilateral  triangle  and  a  square,  each  with  area  a^.  The  side  of 
the  square  is  a,  and  the  perimeter  4  a.  The  area  of  the  equilateral  tri- 
angle is  I s2  Vs.  Therefore  ^ s^  Vs  =  a^^  or  ^  s</d  =  a.  Now  </s  =  VVs ; 
hence  we  have  V3  =  1.734-,and  v  V3=  Vl. 73  =  1.3 +  •  Hence  is  x  1.3  =  a, 
and  s  =  1.5  a,  and  the  perimeter  of  the  triangle  is  4.5  a.  Therefore  the 
perimeter  of  the  square  is  less  than  that  of  the  triangle. 


272  APPENDIX  TO  PLANE  GEOMETRY 

EXERCISE  73 
Maxima  and  Minima 

1.  Of  all  equivalent  parallelograms  that  have  equal  bases, 
the  rectangle  has  the  minimum  perimeter. 

2.  Of  all  equivalent  rectangles,  the  square  has  the  minimum 
perimeter. 

3.  Of  all  triangles  that  have  the  same  base  and  the  same 
altitude,  the  isosceles  has  the  minimum  perimeter. 

4.  Of  all  triangles  that  can  be  inscribed  in  a  given  circle,  the 
equilateral  is  the  maximum  and  has  the  maximum  perimeter. 

5.  To  inscribe  in  a  semicircle  the  maximum  rectangle. 

6.  Find  the  area  of  the  maximum  triangle  inscribed  in  a 
semicircle  whose  radius  is  3  in. 

7.  Of  all  polygons  of  a  given  number  of  sides  that  can 
be  inscribed  in  a  given  circle,  that  which  is  regular  has  the 
maximum  area  and  the  maximum  perimeter. 

8.  Of  all  polygons  of  a  given  number  of  sides  that  can  be 
circumscribed  about  a  given  circle,  that  which  is  regular  has 
the  minimum  area  and  the  minimum  perimeter. 

9.  In  a  given  line  required  to  find  a  point  such  that  the 
sum  of  its  distances  from  two  given  points  on  the  same 
side  of  the  line  shall  be  the  minimum.  a 

How  does  AP  -\-  PB  compare  with  A'B?  and     -^ 
this  with  A'X-^  XB  ?  and  this  with  AX+  XB  ?      J^ 
This  is  the  problem  of  a  ray  of  light  from  -4  to 
the  mirror  CD,  and  reflected  to  B. 

10.  To  divide  a  given   line  into  two  1' 
segments  such  that  the  sum  of  the  squares  on  these  segments 
shall  be  the  minimum. 

11.  To  divide  a  given  line  into  two  segments  such  that  their 
product  shall  be  the  maximum. 


EECREATIONS  273 

421.  Recreations  of  Geometry.  The  following  simple  puzzles 
and  recreations  of  geometry  may  serve  the  double  purpose  of 
adding  interest  to  the  study  of  the  subject  and  of  leading  the 
student  to  exercise  greater  care  in  his  demonstrations.  They 
have  long  been  used  for  this  purpose  and  are  among  the  best 
known  puzzles  of  geometry. 

EXERCISE  74 

1.  To  prove  that  every  triangle  is  isosceles. 

Let  ABC  be  a  A  that  is  not  isosceles. 

Take  CP  the  bisector  of  ZACB,  and  ZP  the  ±  bisector  of  AB. 

These  lines  must  meet,  as  at  P,  for  otherwise  c 

they  would  be  II,  which  would  require  CP  to  be  _L  j/f\ 

to  AB,  and  this  could  only  happen  if  A  ABC  were  Y/    /    \x 

isosceles,  which  is  not  the  case  by  hypothesis.  >:     ^'''''\  -/ 

From  P  draw  PX  ±  to  BC  and  PY  ±  to  C^,  and  ^^^^:'  |  "><^^ 
draw  PA  and  PB.  ''   ^' 

Then  since  ZP  is  the  ±  bisector  of  AB,  .-.  PA  =  PB. 

And  since  CP  is  the  bisector  of  ZACB,  .-.  PX  =  PY. 

.-.  the  rt.  A PBX  and  PA Y  are  congruent,  and  BX  =  AY. 

But  the  rt.  A  PXC  and  PYC  are  also  congruent,  and  .-.  XC  =  YC. 

Adding,  we  have  BX  +  XC  =  AY -\- YC,  or  BC  =  AC. 

.-.  A  ABC  is  isosceles  even  though  constructed  as  not  isosceles. 

2.  To  prove  that  part  of  an  angle  equals  the  whole  angle. 

Take  a  square  ABCD,  and  draw  MM'P,  the  ±  bisector  of  CD.  Then 
Jfilf'P  is  also  the  ±  bisector  of  ylB.  D  M  c 

From  B  draw  any  line  BX  equal  to  AB. 

Draw  BX  and  bisect  it  by  the  _L  NP. 

Since  BX  intersects  CD,  Js  to  these  lines  can- 
not be  parallel,  and  must  meet  as  at  P. 

Draw  PA,  PB,  PC,  PX,  and  PB. 

Since  MP  is  the  ±  bisector  of  CD,  PD=PC. 
Similarly  PA  =  PB,  and  PD  =  PX.  X'fi^'''' 

.-.  PX  =  PD  =  PC.  K 

But  BX  —  BC  by  construction,  and  PB  is  common  to  A  PBX  and  PBC. 

.:  A  PBX  is  congruent  to  A  PBC,  and  Z  XBP  =  Z  CBP. 

.:  the  whole  Z  XBP  equals  the  part,  Z  CBP, 


274 


APPENDIX  TO  PLANE  GEOMETRY 


3.  To  prove  that  part  of  an  angle  equals  the  whole  angle. 

Take  a  right  triangle  ABC,  and  con- 
struct upon  tlie  hypotenuse  BC  an  equi- 
lateral triangle  BCD,  as  shown. 
On  CD  lay  off  CP  equal  to  CA. 
Through  X,  the  mid-point  of  AB, 
draw  PX  to  meet  CB  produced  at  Q. 
Draw  QA. 

Draw  the  ±  bisectors  of  QA  and 
QP,  as  YO  and  ZO.  These  must  meet 
at  some  point  0  because  they  are  ±  to 
two  intersecting  lines. 

Draw  OQ,  OA,  OP,  and  OC. 

Since  0  is  on  the  ±  bisector  of  QA,  .-.  OQ  =  OA. 

Similarly  OQ  =  OP,  and  .-.  OA  =  OP. 

But  CA  =  CP,  by  construction,  and  CO  =  CO. 

.:  AA OC  is  congruent  to  A  POC,  and  ZACO  =  Z PCO. 

4.  To  prove  that  part  of  a  line  equals  the  whole  line. 

Take  a  triangle  ABC,  and  draw  CP  ±  to  ^-B. 
From  C  draw  CX,  making  Z^CX  =  ZB. 
Then  A  ABC  and  ACX  are  similar. 
.-.  AABC:AACX=BC^:CX^. 
Furthermore    A  ABC:  A  ACX  =  AB:  AX. 
.'.  BC"         " 
or  BC^:AB=CX 

But 


A'  P 


CX^  =  AB:AX, 
^:AX. 


BC^  =  AC'^+  AB^ 


and 


CX' 


AC^  +  AX^ 


f;2 


AC'-{-AB"-2AB-AP 


AC^+AX^-2AX 


2AB.AP, 
2AX-AP. 
AP 


AB 


AX 


^2 


AC 
AB 


AC 


^  AB-2AP  =  ——  +  AX-2AP. 

AX 


AC' 
AB 


AX 


ACT 
AX 


AB, 


AC'-AB-AX      AC'-  AB-AX 


AB 


AX 


AB=  AX. 


RECREATIONS 


275 


5.  To  show  geometrically  that  1=0. 

Take  a  square  that  is  8  units  on  a  side,  and  cut  it  into  three  parts, 
-4,  7?,  C,  as  shown  in  the  right-hand  figure.    Fit 
these  parts  together  as  in  the  left-hand  figure. 

Now  tlie  square  is  8  units  on  a  side,  and  therefore 
contains  8  x  8,  or  64,  small  squares,  while  the  rec- 
tangle is  13  units  long  and  5  units  high,  and  there- 
fore contains 
5  X  13,  or  65, 
small  squares. 
But  the  two 
figures  are  each  made  up  of  J.  +  J5+  C 
(Ax.ll), and  therefore  are  equal  (Ax.8). 
64  we  have  1  z=  0  (Ax.  2). 


+  _4__,__(-_. 


65  =  64,  and  by  subtracting 


6.  To  prove  that  any  point  on  a  line  bisects  it. 

Take  any  point  P  on  AB.  q 

On  AB  construct  an  isosceles  A  ABC,  having  /*\ 

AC=BC;  and  draw  PC.  /  \\ 

Then  in  A  APC  and  PBC,  we  have  /      \  \ 

ZA=ZB,  §  74  /  \    \ 

AC=BC,  Const.         /  \      \ 

and  PC  =  PC.  Iden.     "^  J'         ^ 

Three  independent  parts  (that  is,  not  merely  the  three  angles)  of  one 
triangle  are  respectively  equal  to  three  parts  of  the  other,  and  the  tri- 
angles are  congruent ;  therefore  AP  =  BP  (§  67). 

7.  To  prove  that  it  is  possible  to  let  fall  two  perpendiculars 
to  a  line  from  an  external  point. 

Take  two  intersecting  ©  with  centers  0  and  (7. 

Let  one  point  of  intersection  be  P,  and  draw  the  diameters  PA  and  PD. 

Draw  AD  cutting  the  circumferences  at  B 
and  C.    Then  draw  PB  and  PC. 

Since  Z  PC  A  is  inscribed  in  a  semicircle, 
it  is  a  right  angle.  In  the  same  way,  since 
ZDBP  is  inscribed  in  a  semicircle,  it  also  is 
a  right  angle. 

.-.  PB  and  PC  are  both  X  to  AD. 


276 


APPENDIX  TO  PLANE  GEOMETRY 


8.  To  prove  that  if  two  opposite  sides  of  a  quadrilateral  are 
equal  the  figure  is  an  isosceles  tra]oezoid. 

Given  the  quadrilateral  ABCD,  with  BC  =  DA. 
To  prove  that  ^J5  is  |I  to  DC. 

Draw  MO  and  NO,  the  ±  bisectors  of  AB  and 
CD,  to  meet  at  0. 

li  AB  and  DC  are  parallel,  the  proposition  is  already  proved. 

U  AB  and  DC  are  not  parallel,  then  MO  and  NO  will  meet  at  0,  either 
inside  or  outside  the  figure.    Let  0  be  supposed  to  be  inside  the  figure. 

Draw  OA,  OB,  OC,  OD. 
Then  since  OM  is  the  ±  bisector  of  AB,  .-.  OA  =  OB. 


Similarly 


and 

Also, 
and 

Similarly 
and 


OD  =  OC. 

But  DA  is  given  equal  to  BC. 
.'.  AAOD  is  congruent  to  A  BOC, 

ZDOA  =  ZBOC. 
rt.  ii  OCN  and  ODN  are  congruent, 

ZNOD  =  ZCON. 


rt.  A  AMO  and  BMO  are  congruent, 
ZAOM  =  ZMOB. 
.:  Z  NOD  +  Z  DO  A  -\-ZAOM=Z  CON  +  Z  BOC  +  Z  MOB, 
or  Z  NOM  =  Z  MON  =  a  st.  Z. 

Therefore  the  line  MON  is  a  straight  line,  and  hence  AB  is  II  to  DC. 


D  ^^^, 


If  the  point  0  is  outside  the  quadrilateral,  as 
In  the  second  figure,  the  proof  is  substantially  the 
same. 

For  it  can  be  easily  shown  that  A/ 

ZDON-ZDOA-ZAOM  1/ 

=  Z  NOC  -  Z  BOC  -  Z  MOB,  ^ 

which  is  possible  only  if 

ZDON=ZDOM, 
or  if  ON  lies  along  OM. 

But  that  the  proposition  is  not  true  is  evident  from  the 
third  figure,  in  which  BC  =  DA,  but  ^Z^  is  not  II  to  DC. 


O 


'N   \ 


M 


HISTORY  OF  GEOMETRY  277 

422.  History  of  Geometry.  The  geometry  of  very  ancient 
peoples  was  largely  the  mensuration  of  simple  areas  and 
volumes  such  as  is  taught  to  children  in  elementary  arithmetic 
to-day.  They  learned  how  to  find  the  area  of  a  rectangle, 
and  in  the  oldest  mathematical  records  that  we  have  there  is 
some  discussion  of  triangles  and  of  the  volumes  of  solids. 

The  earliest  documents  that  we  have,  relating  to  geometry, 
come  to  us  from  Babylon  and  Egypt.  Those  from  Babylon 
were  written  about  2000  b.c.  on  small  clay  tablets,  some  of 
them  about  the  size  of  the  hand,  these  tablets  afterwards 
having  been  baked  in  the  sun.  They  show  that  the  Baby- 
lonians of  that  period  knew  something  of  land  measures,  and 
perhaps  had  advanced  far  enough  to  compute  the  area  of  a 
trapezoid.  For  the  mensuration  of  the  circle  they  later  used, 
as  did  the  early  Hebrews,  the  value  tt  =  3. 

The  first  definite  knowledge  that  we  have  of  Egyptian  math- 
ematics comes  to  us  from  a  manuscript  copied  on  papyrus,  a 
kind  of  paper  used  about  the  Mediterranean  in  early  times. 
This  copy  was  made  by  one  Aah-mesu  (The  Moon-born),  com- 
monly called  Ahmes,  who  probably  flourished  about  1700  b.c. 
The  original  from  which  he  copied,  written  about  2300  b.c, 
has  been  lost,  but  the  papyrus  of  Ahmes,  written  nearly  four 
thousand  years  ago,  is  still  preserved  and  is  now  in  the  British 
Museum.  In  this  manuscript,  which  is  devoted  chiefly  to  frac- 
tions and  to  a  crude  algebra,  is  found  some  work  on  mensu- 
ration. Among  the  curious  rules  are  the  incorrect  ones  that 
the  area  of  an  isosceles  triangle  equals  half  the  product  of 
the  base  and  one  of  the  equal  sides;  and  that  the  area  of  a 
trapezoid  having  bases  &,  h\  and  nonparallel  sides  each  equal 
to  (X,  is  ^a(b-\-b').  One  noteworthy  advance  appears  however. 
Ahmes  gives  a  rule  for  finding  the  area  of  a  circle,  substan- 
tially as  follows :  Multiply  the  square  on  the  radius  by  (-V-)^ 
which  is  equivalent  to  taking  for  ir  the  value  3.1605.  Long 
before  the  time  of  Ahmes,  however,  Egypt  had  a  good  working 


278  APPENDIX  TO  PLANE  GEOMETRY 

knowledge  of  practical  geometry,  as  witness  the  building  of 
the  pyramids,  the  laying  out  of  temples,  and  the  digging  of 
irrigation  canals. 

From  Egypt  and  possibly  from  Babylon  geometry  passed  to 
the  shores  of  Asia  Minor  and  Greece.  The  scientific  study  of 
the  subject  begins  with  Thales,  one  of  the  Seven  Wise  Men 
of  the  Grecian  civilization.  Born  at  Miletus  about  640  b.c,  he 
died  at  Athens  in  548  b.c.  .  He  spent  his  early  manhood  as  a 
merchant,  accumulating  the  wealth  that  enabled  him  to  spend 
his  later  years  in  study.  He  visited  Egypt  and  is  said  to  have 
learned  such  elements  of  geometry  as  were  known  there.  He 
founded  a  school  of  mathematics  and  philosophy  at  Miletus, 
known  as  the  Ionic  School.  How  elementary  the  knowledge 
of  geometry  then  was,  may  be  understood  from  the  fact  that 
tradition  attributes  only  about  four  propositions  to  Thales, 
substantially  those  given  in  §§  60,  72,  74,  and  215  of  this  book. 

The  greatest  pupil  of  Thales,  and  one  of  the  most  remark- 
able men  of  antiquity,  was  Pythagoras.  Born  probably  on  the 
island  of  Samos,  just  off  the  coast  of  Asia  Minor,  about  the 
year  580  b.c,  Pythagoras  set  forth  as  a  young  man  to  travel. 
He  went  to  Miletus  and  studied  under  Thales,  probably  spent 
several  years  in  study  in  Egypt,  very  likely  went  to  Babylon, 
and  possibly  went  even  to  India,  since  tradition  asserts  this 
and  the  nature  of  his  work  in  mathematics  confirms  it.  In 
later  life  he  went  to  southern  Italy,  and  there,  at  Crotona,  in 
the  southeastern  part  of  the  peninsula,  he  founded  a  school 
and  established  a  secret  society  to  propagate  his  doctrines. 
In  geometry  he  is  said  to  have  been  the  first  to  demonstrate 
the  proposition  that  the  square  on  the  hypotenuse  of  a  right 
triangle  is  equivalent  to  the  sum  of  the  squares  on  the  other 
two  sides  (§  337).  The  proposition  was  known  before  his  time, 
at  any  rate  for  special  cases,  but  he  seems  to  have  been  the 
first  to  prove  it.  To  him  or  to  his  school  seems  also  to  have 
been  due  the  construction  of  the  regular  pentagon  (§§  397, 398) 


HISTORY  OF  GEOMETRY  279 

and  of  the  five  regular  polyhedrons.  The  construction  of  the 
regular  pentagon  requires  the  dividing  of  a  line  in  extreme 
and  mean  ratio  (§  311),  and  this  problem  is  commonly  assigned 
to  the  Pythagoreans,  although  it  played  an  important  part  in 
Plato's  school.  Pythagoras  is  also  said  to  have  known  that  six 
equilateral  triangles,  three  regular  hexagons,  or  four  squares, 
can  be  placed  about  a  point  so  as  just  to  fill  the  360°,  but  that 
no  other  regular  polygons  can  be  so  placed.  To  his  school  is 
also  due  the  proof  that  the  sum  of  the  angles  of  a  triangle 
equals  two  right  angles  (§  107),  and  the  construction  of  at 
least  one  star-polygon,  the  star-pentagon,  which  became  the 
badge  of  his  fraternity. 

For  two  centuries  after  Pythagoras  geometry  passed  through 
a  period  of  discovery  of  propositions.  The  state  of  the  science 
may  be  seen  from  the  fact  that  (Enopides  of  Chios,  who 
flourished  about  465  b.c,  showed  how  to  let  fall  a  perpendicu- 
lar to  a  line  (§  227),  and  how  to  construct  an  angle  equal  to  a 
given  angle  (§  232).  A  few  years  later,  about  440  b.c,  Hippoc- 
rates of  Chios  wrote  the  first  Greek  textbook  on  mathematics. 
He  knew  that  the  areas  of  circles  are  proportional  to  the  squares 
on  their  radii,  but  was  ignorant  of  the  fact  that  equal  central 
angles  or  equal  inscribed  angles  intercept  equal  arcs. 

About  430  B.C.  Antiphon  and  Bryson,  two  Greek  teachers, 
worked  on  the  mensuration  of  the  circle.  The  former  attempted 
to  find  the  area  by  doubling  the  number  of  sides  of  a  regular 
inscribed  polygon,  and  the  latter  by  doing  the  same  for  both  in- 
scribed and  circumscribed  polygons.  They  thus  substantially 
exhausted  the  area  between  the  circle  and  the  polygon,  and 
hence  this  method  was  known  as  the  Method  of  Exhaustions. 

During  this  period  the  great  philosophic  school  of  Plato 
(429-348  B.C.)  flourished  at  Athens,  and  to  this  school  is  due 
the  first  systematic  attempt  to  create  exact  definitions,  axioms, 
and  postulates,  and  to  distinguish  between  elementary  and 
higher  geometry.    At  this  time  elementary  geometry  became 


280  APPENDIX  TO  PLANE  GEOMETRY 

limited  to  the  use  of  the  compasses  and  the  unmarked  straight 
edge,  which  took  from  this  domain  the  possibility  of  con- 
structing a  square  equivalent  to  a  given  circle  ("  squaring  the 
circle"),  of  trisecting  any  given  angle,  and  of  constructing 
a  cube  with  twice  the  volume  of  a  given  cube  ("  duplicating 
the  cube"),  these  being  the  three  most  famous  problems  of 
antiquity.  Plato  and  his  school  were  interested  in  the  so-called 
Pythagorean  numbers,  numbers  that  represent  the  three  sides 
of  a  right  triangle.  Pythagoras  had  already  given  a  rule  to 
the  effect  that  i(m''-{-lf  =  7n^  +  i(m^-iy.  The  school  of 
Plato  found  that  [(^my-\-iy  =  m^-i-[(^my-lf.  By  giving 
various  values  to  vi,  different  numbers  will  be  found  such  that 
the  sum  of  the  squares  of  two  of  them  is  equal  to  the  square  of 
the  third. 

The  first  great  textbook  on  geometry,  and  the  most  famous 
one  that  has  ever  appeared,  was  written  by  Euclid,  who  taught 
mathematics  in  the  great  university  at  Alexandria,  Egypt, 
about  300  B.C.  Alexandria  was  then  practically  a  Greek  city, 
having  been  named  in  honor  of  Alexander  the  Great,  and 
being  ruled  by  the  Greeks. 

Euclid's  work  is  known  as  the  "Elements,"  and,  as  was  the  case 
with  all  ancient  works,  the  leading  divisions  were  called  books, 
as  is  seen  in  the  Bible  and  in  such  Latin  writers  as  Caesar 
and  Vergil.  This  is  why  we  speak  of  the  various  books  of 
geometry  to-day.  In  this  work  Euclid  placed  all  the  leading 
propositions  of  plane  geometry  as  then  known,  and  arranged 
them  in  a  logical  order.  Most  subsequent  geometries  of  any  im- 
portance since  his  time  have  been  based  upon  Euclid,  improving 
the  sequence,  symbols,  and  wording  as  occasion  demanded. 

Euclid  did  not  give  much  solid  geometry  because  not  much 
was  known  then.  It  was  to  Archimedes  (287-212  b.c),  a 
famous  mathematician  of  Syracuse,  on  the  island  of  Sicily, 
that  some  of  the  most  important  propositions  of  solid  geometry 
are  due,  particularly  those  relating  to  the  sphere  and  cylinder. 


HISTORY  OF  GEOMETRY  281 

He  also  showed  how  to  find  the  approximate  value  of  tt  by  a 
method  similar  to  the  one  we  teach  to-day  (§  404),  proving 
that  the  real  value  lies  between  3}  and  3}^.  Tradition  says 
that  the  sphere  and  cylinder  were  engraved  upon  his  tomb. 
The  Greeks  contributed  little  more  to  elementary  geometry, 
although  Apollonius  of  Perga,  who  taught  at  Alexandria  be- 
tween 250  and  200  b.c,  wrote  extensively  on  conic  sections;  and 
Heron  of  Alexandria,  about  the  beginning  of  the  Christian  era, 
showed  that  the  area  of  a  triangle  whose  sides  are  a,  b,  c,  equals 
V6'(.s-  —  a) (s  —  b) (s  —  c),  where  s  =  ^(a-\-b^-c)  (see  p.  211). 

The  East  did  little  for  geometry,  although  contributing 
considerably  to  algebra.  The  first  great  Hindu  writer  was 
Aryabhatta,  who  was  born  in  476  a.d.  He  gave  the  very 
close  approximation  for  tt,  expressed  in  modern  notation  as 
3.1416.  The  Arabs,  about  the  time  of  the  Arabian  Nights  Tales 
(800  A.D.),  did  much  for  mathematics,  translating  the  Greek 
authors  into  their  own  language  and  also  bringing  learning 
from  India.  Indeed,  it  is  to  them  that  modern  Europe  owes 
its  first  knowledge  of  Euclid.  They  contributed  nothing  of 
importance  to  geometry,  however. 

Euclid  was  translated  from  the  Arabic  into  Latin  in  the 
twelfth  century,  Greek  manuscripts  not  being  then  at  hand,  or 
being  neglected  because  of  ignorance  of  the  language.  The 
leading  translators  were  Athelhard  of  Bath  (1120),  an  English 
monk  who  had  learned  Arabic  in  Spain  or  in  Egypt ;  Gerhard 
of  Cremona,  an  Italian  monk ;  and  Johannes  Campanus,  chap- 
lain to  Pope  Urban  IV. 

In  the  Middle  Ages  in  Europe  nothing  worthy  of  note  was 
added  to  the  geometry  of  the  Greeks.  The  first  edition  of 
Euclid  was  printed  in  Latin  in  1482,  the  first  one  in  English 
appearing  in  1570.  Our  symbols  are  modern,  -|-  and  —  first 
appearing  in  a  German  work  in  1489;  =  in  Recorde's  "Whet- 
stone of  Witte"  in  1557;  >  and  <  in  the  works  of  Harriot 
(1560-1621);  and  x  in  a  publication  by  Oughtred  (1574-1660). 


282  APPENDIX  TO  PLANE  GEOMETRY 

423.  Notation  used  in  Formulas.  Following  the  general  cus- 
tom, small  letters  represent  numerical  values,  large  letters  rep- 
resent points.  The  following  abbreviations  have  been  used  in 
this  book  as  consistently  as  the  circumstances  would  allow,  the 
context  telling  which  abbreviation  is  intended : 

a  =  area,  apothem  I  =  length. 

a,  b,  c  =  sides  oiAABC.  m  =  median, 
a' =  projection  of  a.  2^  —  perimeter. 

b,  h'  =  bases.  ^  r  =  radius. 

c  =  circumference.  s  =  semiperimeter  of  A, 

d  =  diameter,  diagonal.  1  (a  +  Z»  -f  c). 

h  =  height,  altitude.  ir  =  3.1416,  or  about  3}. 

424.  Formulas  for  Line  Values.  The  following  are  the  most 
important  formulas  in  line  values : 

Right  triangle,  a'  -i-b''  =  c"  (§  337). 

Any  triangle,  a^ -^  b'' ±  2  aV  =  c"  {^%  341,  342). 

Circle,  c  =  2  irr  =  ird  (§  385). 

Radius  of  circle,  v  —  c-^2'Tr. 

Equilateral  triangle,  ^^  =  \b  y 3. 

Diagonal  of  square,  d  =  b  V2  (§  339). 

Side  of  square,  b  =  Va. 

425.  Areas  of  Plane  Figures.  The  following  are  the  formulas 
for  the  areas  of  the  most  important  j^lane  figures : 

Rectangle,  bh  (§  320). 

Square,  b''  (§  320). 

Parallelogram,  bh  (§  322). 


Triangle,  ^^bh(^S25),^s(s-a)(s-b){s-c). 

Equilateral  triangle,  ^  b'^  v  3. 

Trapezoid,  V^(^  +  ^')  (§^29). 

Regular  polygon,  ^  ap  (§  386). 

Circle,  ^-rc  =  7rr2(§§388,  389). 


INDEX 


Page 

Acute  angle 16 

Acute  triangfe 26 

Adjacent  angles 7 

Alternation,  proportion  by      .152 

Altitude 59 

Analytic  proof      .     .    80,  140,  141 

Angle 6 

acute 16 

at  center  of  regular  polygon  227 

central 93 

complement  of      ....     18 

conjugate  of 18 

exterior 51 

inscribed 115 

measure  of 18 

oblique 16 

obtuse 16 

plane 6 

reentrant     .     .     .     ^     .     .     08 

reflex 16 

right 7,  16 

sides  of 6 

size  of 6,  17 

straight 16 

supplement  of 18 

vertex  of 6 

Angles,  adjacent 7 

alternate-interior      .     .     .     47 
complementary    ....     18 

conjugate 18 

corresponding 26 

equal 6 


Page 

Angles,  exterior    .     .     ,     .    47,  51 

exterior-interior  ....     47 

generation  of 17 

interior 47,  51 

made  by  a  transversal  .     .     47 

of  a  polygon 68 

of  a  triangle 7 

supplementary     .     .     .     .     18 

vertical 18 

Antecedents 151 

Apothem 227 

Arc 7,93 

major 93 

minor 93 

Area  of  circle 115 

of  irregular  polygon      .     .  199 

of  surface 191 

Attack,  methods  of   .     .     140,  145 

Axiom 21 

Axioms,  list  of 22 

Axis  of  symmetry     ....  261 

Base .    7, 32,  59 

Bisector 6,  74 

Broken  line 5 

Center  of  circle 7 

of  regular  polygon    .     .     .  227 

of  symmetry 261 

Central  angle 93 

Chord 95 

Circle 7,93 


283 


284 


INDEX 


Page 

Circle,  arc  of    .     .     . 

.     .      7,  93 

area  of    ...     . 

.     .     .115 

as  a  limit     .     .     . 

.     114,  237 

as  a  locus    .     .     . 

...     93 

center  of     .     .     . 

...       7 

central  angle  of    . 

...     93 

chord  of      .     .     . 

...     95 

circumference  of  . 

...       7 

circumscribed  .     . 

.     .     .114 

diameter  of      .     . 

.     .      7,  93 

inscribed     .     .     . 

...  114 

radius  of      .     .     . 

.     .      7,  93 

secant  to     .     .     , 

.     102,  177 

sector  of      .     .     . 

.     .     .  115 

segment  of  .     .     . 

.     .     .115 

tangent  to  .     .     . 

.     .     .102 

Circles,  concentric     . 

.     .     .104 

escribed .... 

...  137 

tangent  .... 

.     .     .107 

Circumcenter  .     .     . 

.     .  78,  136 

Circumference      .     . 

...       7 

Circumscribed  circle 

.     .     .114 

Circumscribed  polygor 

1 .     .     .114 

Commensurable  magn 

tudes    .  112 

Common  measure 

.     .     .112 

Common  tangents 

.     .     .109 

Complement     .     .     . 

...     18 

Composition,  proportic 

m  by     .153 

Concave  polygon  .     . 

...     68 

Concentric  circles      . 

...  104 

Concurrent  lines  .     . 

....     77 

Congruent   .... 

.     .    26,68 

Conjugate    .... 

...     18 

Consequents     .     .     . 

.     .     .151 

Constant      .... 

.     .     .114 

Continued  proportion 

.     .     .151 

Continuity,  principle  c 

)f      .     .  125 

Converse  propositions 

.     .    35,  95 

Converse  theorems,  la 

w  of      .95 

Convex  polygon    .     . 

.     ,     .     68 

Page 

Corollary 21 

Corresponding  angles  ...  26 
Corresponding  lines  ....  165 
Corresponding  sides  .     .       26,  165 

Curve 5 

Curvilinear  figure      ....       6 

Decagon 68 

Degree  ....  .^  ...  18 
Determinate  cases     ....  140 

Diagonal 59,  68 

Diameter 7,  93 

Difference  of  magnitudes  .     .     17 

Dimensions 2 

Discussion  of  a  problem      126,  140 

Distance 42 

Division,  harmonic    ....  161 

proportion  by 154 

Dodecagon  .......     68 

Drawing  figures    .     .     .     8,  29,  84 

Equal  angles 6 

Equal  lines 5 

Equiangular  polygon      ...  68 

Equiangular  triangle      ...  26 

Equilateral  polygon  ....  68 

Equilateral  triangle  ....  26 

Equivalent  figures     .     .     .     .  191 

Escribed  circles 137 

Excenter 137 

Exterior  angles     .     .     .     .    47,  51 

Extreme  and  mean  ratio    .     .  184 

Extremes     .......  151 

Figure 4 

curvilinear 5 

geometric    ......  4 

plane 4 

rectilinear 6 

symmetric 261 


INDEX 


285 


Figures,  equivalent   . 

isoperimetric   .     . 

symmetric  .  .  . 
Foot  of  perpendicular 
Formulas  .... 
Fourth  proportional  . 

Generation  of  angles 
of  magnitudes .     . 

Geometric  figure  .     . 

Geometry     .     .     .     .' 
History  of  .     .     . 


Harmonic  division  . 

Heptagon     .     .     .  . 

Hexagon      .     .     .  . 
History  of  Geometry 

Homologous  angles  . 
Homologous  lines 
Homologous  sides 

Hypotenuse      .     .  . 

Hypothesis  .     .     .  . 


Impossible  cases 
Incenter .     .     . 
Incommensurable  magnitudes 
Incommensurable  ratio 
Indeterminate  cases  . 
Indirect  proof  . 
Inscribed  angle     .     . 
Inscribed  circle     . 
Inscribed  polygon 
Instruments 
Interior  angles      .     . 
Inversion,  proportion 
Isoperimetric  polygons 
Isosceles  trapezoid    . 
Isosceles  triangle  .     . 


by 


Limit 


Page  page 

.  191       Limits,  principle  of   ...     .  115 

.  265      Line 3,5 

.  261  broken 5 

7  curve      5 

.  282  of  centers 107 

.  151  segments  of      ....    5,  161 

straight 5 

.     17       Lines,  concurrent      .     .     .     .77 
4,17  corresponding.     ,     .•    .     ,  165 

4  equal 5 

4  oblique 16 

.  277  parallel 46 

perpendicular 7 

product  of 194 

transversal  of 47 

Loci,  solutions  by      ....  143 

Locus 73 

proof  of 74 

Magnitudes 3 

bisectors  of 6 

commensurable    ....  112 

constant 114 

differences  of 17 

generation  of  .     .     .     .     4,  17 
incommensurable      .     .     .112 

sums  of 17 

variable 114 

Maximum 265 

Mean  proportional     ....  151 

Means 151 

Measure 112 

.       8  angle 18 

47,  51  common 112 

.  153  numerical    ....     112,  117 

.  265      Median 77 

.     59       Methods  of  attack      .     .     140,  145 
.     26  of  proof.     .       35,77,80,8.3,84 

Minimum 265 

114,  237      Multiple 112 


161 


277 
26 

165 
26 
42 
30 

140 
137 
112 
113 
140 
83 
115 
114 
114 


286 


INDEX 


Page 

Nature  of  proof 25 

of  solution 126 

Negative  quantities   ....  125 

Nonagon 68 

Numerical  measure   .     .     112,  117 

Oblique  angle 16 

Oblique  lines 16 

Obtuse  angle 16 

Obtuse  triangle 26 

Octagon .     .  68 

Optical  illusions 15 

Parallel  lines 46 

Parallelogram 59 

Pentagon 68 

Pentadecagon 246 

Perigon 18 

Perimeter 7,  68 

Perpendicular 7 

Perpendicular  bisector  ...     74 

Pi(7r) 238 

value  of 249 

Plane 3 

angle 6 

geometry 4 

Point 3 

of  contact    ....     102, 107 

Polygon 68 

angles  of 68 

apothem  of  regular  .     .     .227 

area  of 191,  199 

center  of  regular  ....  227 
circumcenter  of    ...     .  136 

circumscribed 114 

concave 68 

convex 68 

diagonal  of      ....    59,  68 

equiangular 68 

equilateral 68 


Page 

Polygon,  incenter  of      .     .     .  137 

inscribed 114 

perimeter  of 68 

radius  of  regular  ....  227 

regular .  68,  227 

sides  of 68 

vertices  of 68 

Polygons,  classified    .     .       68,  114 

congruent 68 

isoperimetric 265 

mutually  equiangular    .     .     68 
mutually  equilateral      .     .     68 

similar 165 

Positive  quantities     ....  125 

Postulate .     21 

of  parallels 46 

Postulates,  list  of       ....     23 
Principle  of  continuity  .     .     .  125 

of  limits 115 

Problem 21,  126 

how  to  attack  .     .     .     140,  145 

Product  of  lines 194 

Projection 205 

Proof,  methods  of  35,  77,  80,  83,  84 

nature  of 25 

necessity  for 15 

Proportion 151 

continued 151 

nature  of  quantities  in  a    .155 
Proportional,  fourth .     .     .     .151 

mean 151 

reciprocally 177 

third 151 

Proposition 21 

Pythagorean  theorem     .     .     .  204 

Quadrilateral 59,  68 

Quadrilaterals  classified     .     .     59 

Radius 7,  93 

of  regular  polygon    .     .     .  227 


INDEX 


287 


Page 
Ratio 112 

extreme  and  mean    .     .     .184 
incommensurable      .     .     .113 

of  similitude 165 

Recreations  of  Geometry    .     .  273 

Rectangle 59 

Rectilinear  figure 5 

Reductio  ad  absurdum  ...     83 

Reentrant  angle 68 

Reflex  angle 16 

Regular  polygon    .     .     .     .68,  227 

Rhomboid 59 

Rhombus 59 

Right  angle 7,  16 

Right  triangle 26 

Scalene  triangle 26 

Secant 102,  177 

Sector 115 

Segment  of  a  circle    .     .     .     .115 

of  a  line 5,  161 

Semicircle 93,  115 

Sides,  corresponding      ...     26 

of  angle 6 

of  triangle 7 

of  polygon  .     .     ...     .     68 

Similar  parts  of  circles  .     .     .  239 

Similar  polygons 165 

Similitude,  ratio  of    ...     .  165 

Size  of  angle 6,  17 

Solid 2 

Solid  geometry 4 

Solution,  nature  of    ...     .  126 

Square 26 

Straight  angle 16 

Straight  line 5 

Subtend 93,  95 

Sum  of  magnitudes   .     .     .     .     17 

Superposition 35 

Supplement 18 


Page 
Surface 3,  191 

Symmetric  figures     ....  261 

Symmetry 261 

Synthetic  proof     .     .      35,  77,  140 

Tangent  ....       102,  107,  109 

Tangent  circles 107 

Terms  of  a  proportion    .     .     .151 

Theorem .     21 

Third  proportional     .     .     .     .151 

Transversal 47 

Trapezium 59 

Trapezoid 59 

Triangle 7,  68 

acute 26 

altitude  of 59 

angles  of 7 

base  of 7,  59 

circumcenter  of    .     .     .  78,  136 

equiangular 26 

equilateral 26 

excenter  of 137 

incenter  of 78,  137 

isosceles 26 

obtuse 26 

right 26 

scalene 26 

sides  of 7 

vertices  of 7 

Triangles  classified    ....     26 

Unit  of  measure 112 

of  surface 191 

Variable 114 

Vertex  of  angle 6 

of  isosceles  triangle        .     .  59 

Vertical  angles 18 

Vertices  of  a  polygon     ...  68 

of  a  triangle 7 


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